Complex Analysis

Xue-Mei Li

May 16, 2016

Contents

1 Complex Differentiation 61.1 The Complex Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Complex Functions of One Variables . . . . . . . . . . . . . . . . . . 61.3 Complex Linear Functions . . . . . . . . . . . . . . . . . . . . . . . 101.4 Complex Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 101.5 The ∂ and ∂ operator . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.7 Holomorphic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 141.8 Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 The Riemann Sphere and Mobius Transforms 162.1 Conformal Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Mobius Transforms (Lecture 5) . . . . . . . . . . . . . . . . . . . . . 17

2.2.1 The Extended Complex Plane . . . . . . . . . . . . . . . . . 182.2.2 Properties of Mobius Transforms . . . . . . . . . . . . . . . 18

2.3 The Riemann Sphere and Stereographic Projection (lecture 7) . . . . . 22

3 Power Series 253.1 Power series is holomorphic in its disc of convergence . . . . . . . . 253.2 Analytic Continuation (Lecture 8) . . . . . . . . . . . . . . . . . . . 273.3 The Exponential and Trigonometric Functions . . . . . . . . . . . . . 293.4 The Logarithmic Function and Power Function (Lecture 9) . . . . . . 30

4 Complex Integration 314.1 Complex Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 Integration Along a Curve (lecture 9) . . . . . . . . . . . . . . . . . . 324.3 Existence of Primitives (Lecture 9) . . . . . . . . . . . . . . . . . . . 344.4 Goursat’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.5 Cauchy’s Theorem: simply connected domains (Lecture 12) . . . . . 394.6 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Cauchy’s Integral Formula 425.1 Keyhole Operation and Other Techniques (Lecture 12) . . . . . . . . 425.2 Cauchy’s Integral Formula . . . . . . . . . . . . . . . . . . . . . . . 445.3 Taylor Expansion, Cauchy’s Derivative Formulas . . . . . . . . . . . 455.4 Estimates, Liouville’s Thm and Morera’s Thm . . . . . . . . . . . . . 46

5.4.1 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . 485.5 Locally Uniform Convergent Sequence of Holomorphic Functions . . 48

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5.6 Schwartz Reflection Principle (Lecture 15) . . . . . . . . . . . . . . . 495.7 The Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . 515.8 Zero’s of Analytic Functions . . . . . . . . . . . . . . . . . . . . . . 525.9 Uniqueness of Analytic Continuation (Lecture 17) . . . . . . . . . . 555.10 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6 Laurent Series and Singularities 566.1 Laurent Series Development (Lecture 17-18) . . . . . . . . . . . . . 566.2 Classification of Isolated Singularities (Lecture 19) . . . . . . . . . . 61

6.2.1 Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626.2.2 Essential Singularities . . . . . . . . . . . . . . . . . . . . . 64

6.3 Meromorphic Function . . . . . . . . . . . . . . . . . . . . . . . . . 646.3.1 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . 64

7 Winding Numbers and the Residue Theorem 667.1 The Index of a Closed Curve (Lecture 22) . . . . . . . . . . . . . . . 67

7.1.1 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . 697.2 The Residue Theorem (Lecture 23) . . . . . . . . . . . . . . . . . . . 717.3 Compute Real Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 72

8 Fundamental Theorems 748.1 The Argument Principle (Lecture 24) . . . . . . . . . . . . . . . . . . 748.2 Rouche’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

8.2.1 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . 768.3 The Open mapping Theorem (Lecture 25) . . . . . . . . . . . . . . . 76

8.3.1 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . 788.4 Bi-holomorphic Maps on the Disc (lecture 25-26) . . . . . . . . . . . 79

9 The Riemann Mapping Theorem 819.1 Hurwitz’s Theorem (lecture 26) . . . . . . . . . . . . . . . . . . . . . 819.2 Family of Holomorphic Functions (Lecture 28) . . . . . . . . . . . . 829.3 The Riemann Mapping Theorem (Lecture 28-29) . . . . . . . . . . . 849.4 Supplementary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

10 Special Functions 8710.1 Constructing Holomorphic Functions by Integration (Lecture 30) . . . 8710.2 The Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . 8710.3 The zeta function(Lecture 30) . . . . . . . . . . . . . . . . . . . . . 89

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Prologue

This is the lecture notes for the third year undergraduate module: MA3B8. If you neednot be motivated, skip this section.

Complex Analysis is concerned with the study of complex number valued functionswith complex number as domain. Let f : C→ C be such a function. What can we sayabout it? Where do we use such an analysis?

The complex number i =√−1 appears in Fourier Transform, an important tool in

analysis and engineering, and in the Schrodinger equation,

i~∂ψ

∂t= − ~2

2m

∂2ψ

∂2x+ V (x)ψ(t, x),

a fundamental equation of physics, that describes how a wave function of a physicalsystem evolves. Complex Differentiation is a very important concept, this is allured toby the fact that a number of terminologies are associated with ‘complex differentiable’.A function, complex differentiable on its domain, has two other names: a holomor-phic map and an analytic function, reflecting the original approach. The first meantthe function is complex differentiable at every point, and the latter refers to functionswith a power series expansion at every point. The beauty is that the two concepts areequivalent. A complex valued function defined on the whole complex domain is anentire function. Quotients of entire functions are Meromorphic functions on the wholeplane.

A map is conformal at a point if it preserves the angle between two tangent vec-tors at that point. A complex differentiable function is conformal at any point whereits derivative does not vanish. Bi-holomorphic functions, a bi-jective holomorphicfunction between two regions, are conformal in the sense they preserve angles. Of-ten by conformal maps people mean bi-holomorphic maps. Conformal maps are thebuilding blocks in Conformal Field Theory. It is conjectured that 2D statistical mod-els at criticality are conformal invariant. An exciting development is SLE, evolvedfrom the Loewner differential equation describing evolutions of conformal maps. TheSchramann-Loewner Evolution (also known as Stochastic Loewner Evolution, abbre-viated as SLE ) has been identified to describe the limits of a number of lattice modelsin statistical mechanics. Two mathematicians, W. Werner and S. Smirnov, have beenawarded the Fields medals for their works on and related to SLE.

Complex valued functions are built into the definition for Fourier transforms. Forf : R→ R,

f(k) =1√2π

∫ ∞∞

e−ikxdx, k ∈ R.

Fourier transform extends the concept of Fourier series for period functions, is an im-portant tool in analysis and in image and sound processing, and is widely used in elec-trical engineering.

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A well known function in number theory is the Riemann zeta-function,

ζ(s) =

∞∑n=1

1

ns.

The interests in the Riemann-zeta function began with Euler who discovered that theRiemann zeta function can be related to the study of prime numbers.

ζ(s) = Π1

1− p−s.

The product on the right hand side is over all prime numbers:

Π1

1− p−s=

1

1− 2−s· 1

1− 3−s· 1

1− 5−s· 1

1− 7−s· 1

1− 11−s. . .

1

1− p−s. . . .

The Riemann-zeta function is clearly well defined for s > 1 and extends to all complexnumbers except s = 1, a procedure known as the analytic /meromorphic continuationof a real analytic function. Riemann was interested in the following question: howmany prime number are below a given number x? Denote this number π(x). Riemannfound an explicit formula for π(x) in his 1859 paper in terms of a sum over the zerosof ζ. The Riemann hypothesis states that all non-trivial zeros of the Riemann zetafunction lie on the critical line s = 1

2 . The Clay institute in Canada has offered a prizeof 1 million dollars for solving this problem.

In symplectic geometry, symplectic manifolds are often studied together with acomplex structure. The space C is a role model for symplectic manifold. A 2-dimensionalsymplectic manifold is a space that looks locally like a piece of R2 and has a symplecticform, which we do not define here. We may impose in addition a complex structure Jxat each point of x ∈M . The complex structure Jx is essentially a matrix s.t. −J2

x is theidentity and defines a complex structure and leads to the concept of Khaler manifolds.

Finally we should mention that complex analysis is an important tool in combina-torial enumeration problems: analysis of analytic or meromorphic generating functionsprovides means for estimating the coefficients of its series expansions and estimates forthe size of discrete structures.

TopicsHolomorphic Functions, meromorphic functions, poles, zeros, winding numbers (rota-tion number/index) of a closed curve, closed curves homologous to zero, closed curveshomotopic to zero, classification of isolated singularities, analytical continuation, Con-formal mappings, Riemann spheres, special functions and maps. Main Theorems:Goursat’s theorem, Cauchy’s theorem, Cauchy’s derivative formulas, Cauchy’s inte-gral formula for curves homologous to zero, Weirerstrass Theorem, The Argumentprinciple, Rouche’s theorem, Open Mapping Theorem, Maximum modulus principle,Schwartz’s lemma, Mantel’s Theorem, Hurwitz’s theorem, and the Riemann MappingTheorem.

References• L. V. Ahlfors, Complex Analysis, Third Edition, Mc Graw-Hill, Inc. (1979)

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• J. B. Conway. Functions of one complex variables.

• T. Gamelin. Complex Analysis, Springer. (2001)

• E. Hairer, G. Wanner, Analyse Complexe et Series de Fourier.http://www.unige.ch/hairer/poly_complexe/complexe.pdf

• G. J. O. Jameson. A First course on complex functions. Chapman and Hall,(1970).

• E. M. Stein and R. Shakarchi. Complex Analysis. Princeton University Press.(2003)

Acknowledgement. I would like to thank E. Hairer and G. Wanner for the figures inthis note.

5

Chapter 1

Complex Differentiation

1.1 The Complex PlaneThe complex plane C = {x+ iy : x, y ∈ R} is a field with addition and multiplication,on which is also defined the complex conjugation x+ iy = x− iy and modulus (alsocalled absolute value) |z| =

√zz =

√x2 + y2. It is a vector space over R and over C

with the norm |z1 − z2|.We will frequently treat C as a metric space, with distance d(z1, z2) = |z1 − z2|,

and so we understand that a sequence of complex numbers zn converges to a complexnumber z is meant by that the distance |zn − z| converges to zero. The space C withthe above mentioned distance is a complete metric space and so a sequence convergesif and only if it is a Cauchy sequence. Since

|zn − z|2 = |Re(zn)− Re(z)|2 + |Im(zn)− Im(z)|2,

zn converges to z if and only if the real parts of (zn) converge to the real part of z andthe imaginary parts of (zn) converge to the imaginary part of z.

In polar Coordinates z ∈ C can be written as z = reiθ where r = |z| and θ is a realnumber, called the argument. We note specially Euler’s formula:

eiθ = cos(θ) + i sin(θ),

so arg z is a multi valued function. It is standard to take the principal value −π <Argz ≤ π, a rather arbitrary choice.

Since e2πik = 1 for k an integer, the nth root function is multi-valued. If

ωk = e2πkn i, k = 0, 1, . . . , n− 1,

the nth roots of the unity, then

(reiθ)1n = r

1n ei

θnωk.

1.2 Complex Functions of One VariablesTo discuss complex differentiation of a function, we request that it is defined on a subsetof the complex plane C which is open. By a set we would usually mean a subset of the

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complex plane C. A set U is open if about every point in U there is a disc containedentirely in U . We further assume that the set is connected, otherwise we could treat itas a separate function on each connected subset.

A subset of C is connected if any two points from the subset can be connected by acontinuous curve which lies entirely within the subset.

An open set is connected if and only if it is not disconnected in the sense that it isnot the union of two disjoint open sets.

Definition 1.2.1 By a region we mean a connected open subset of C. By a properregion we mean an open connected subset of C that is not the whole complex plane.

From now on, by a function we mean a function f : U → C where U is a region.By an open disc we mean {z : |z − z0| < r} where z0 ∈ C and r > 0. A closed

disc is {z : |z−z0| ≤ r}where z0 ∈ C and r ≥ 0. The unit disc centred at 0 is denotedby

D = {z : |z| < 1}.

Other frequently seen open sets are the deleted discs and the annulus

{z : 0 < |z − z0| < r}, {z : r1 < |z − z0| < r2},

and polygons.

Example 1.2.1 Given z0 ∈ C, the function f : C → C given by the formula f(z) =z + z0 is said to be a translation.

As a set we may wish to identify a complex number s + it with the pair of realnumbers (s, t), so C is identified with R2. Since, for z = x+ iy and c = s+ it,

c(x+ iy) = (sx− ty) + i(tx+ sy),

the map z → cz is represented by a a linear map:(xy

)→(s −tt s

)(xy

).

Multiplication by i is the same as multiply by J on the left, where

J =

(0 −11 0

).

Example 1.2.2 Given c ∈ C, the function f(z) = cz is of the form below. For z =x+ iy, c = |c| eiθ, (

xy

)7→ |c|

(cos(θ) − sin(θ)sin(θ) cos(θ)

)(xy

).

This is the composition of a rotation by an angle θ and a scaling by |c|. This mappreserves the angle between two vectors, i.e. it is a conformal map.

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−1 1

−1

1

−1 1

−1

1

zc

w = c z

c

Figure 1.1: Graph by E. Hairer and G. Wanner

Example 1.2.3 Define f(z) = z2 whose maximal domain of definition is C. Writef = u+ iv. Then

u(x, y) = x2 − y2, v(x, y) = 2xy.

It is easy to see that f takes the horizontal lines y = b where b 6= 0 to parabolas onthe w plane facing right. Solve the equations: x2 − b2 = u and v = 2xb to see

u =1

4b2v2 − b2.

Also, f takes the vertical lines x = a where a 6= 0 to parabolas on the w plane facingleft.

u = 4a2 − 1

4a2v2.

These two sets of parabolas intersect at right angles, see Figure 1.2.If b = 0, the line y = 0 is mapped to the right half of the real axis; the line x = 0 is

mapped to the left half of the real axis. We observe that at 0, these two curves, imagesof the real and complex line from the domain space, fail to intersects with each otherat a right angle. (0 is the only point at which f ′ = 0, explaining the orthogonality andfailing of the orthogonality where the image curves meet, to which we return later.)

The function f(z) = z2 is not injective. It takes the line y = b and y = −b to thesame image. To see this better let us use polar coordinates. Then f(reiθ) = r2e2iθ.When restricted to the positive half complex plane, f is injective. In fact,

f : C+ → C \ [0,∞)

is a bijection with inverse f−1(w) =√w. When restricted to the negative half plane

f : C− → C \ [0,∞),

f is also injective with inverse f−1(w) = ω2√w = −

√w.

8

−1 0

−1

1

−1 1

−1

1

z =√w

w = z2

Figure 1.2: Graph by E. Hairer and G. Wanner

Example 1.2.4 Define f on C \ (−∞, 0] by f(w) =√w, the principal brach of the

square root function. So f(reiθ) =√reiθ/2, −π < θ < π. It has another formula:

f(w) =√|w|ei(Argw/2), w ∈ C \ (−∞, 0].

It maps the slit w plane into the right half of the z-plane. The other branch of thesquare root is −

√w. It is possible to glue the two slit domains together to form a

complex manifold, known as a Riemann surface, so in one sheet (chart) the functiontakes the value of one brach and in the other we use the other brach in a way f changescontinuously as w changes.

Example 1.2.5 The map f(z) = 1z , the inversion map, is defined on C\{0}. It is easy

to see that f takes circles centred at the origin to circles centres at the origin. It take thelocus of the solutions of |z − z0| = r to that of a circle in the w-plane, see Proposition2.2.6 and Example sheets.

Example 1.2.6 (Mobius Transforms) Let a, b, c, d ∈ C where ad 6= bc. Define

f(z) =az + b

cz + d.

If c = 0 the domain of f is C otherwise it is C \ {−dc}.

Exercise 1.2.7 1. Prove that for any real number r, not 1, the equation

|z − z1| = r|z − z2|

determines a circle.

2. Prove that any Mobius transform is a composition of translations, scalings, andinversions.

9

Later we will see that Mobius Transforms can be considered as maps on the extendedcomplex plane, the Riemann sphere.

1.3 Complex Linear FunctionsWe identify R2 with C. A function T : R2 → R2 is real linear if for all z1, z2, z ∈ R2,

T (z1 + z2) = T (z1) + T (z2),

T (rz) = rT (z), ∀r ∈ R

A map T : C→ C is complex linear if for all z1, z2, z ∈ C,

T (z1 + z2) = T (z1) + T (z2),

T (kz) = kT (z), ∀k ∈ C

Proposition 1.3.1 A real linear function T : R2 → R2 is complex linear iff

T (i) = iT (1).

We now look at the matrix representations. Every real linear map is of the form(xy

)→(a bc d

)(xy

).

If k = s+ it, the complex linear map T (z) = kz is given by

T

(xy

)=

(s −tt s

)(xy

).

For every real linear map T there exists a unique pair of complex numbers λ and µsuch that

T (z) = λz + µz,

which is complex linear if and only if µ = 0. Furthermore,

λ =1

2

((a+ ic) +

1

i(b+ id)

), µ =

1

2

((a+ ic)− 1

i(b+ id)

)

1.4 Complex DifferentiationLet f = u+ iv, defined in a region U . When C is identified as R2 we may treat u andv as real valued functions on R2. In this way f is an R2 valued function of two realvariables x and y. Then f is (real) differentiable at (x0, y0) if there exists a linear map(df)(x0,y0) : R2 → R2 and a function φ such that

f(x, y) = f(x0, y0) + (df)(x0,y0)

(x− x0

y − y0

)+ φ(x, y)

∣∣∣∣(x− x0

y − y0

)∣∣∣∣ (1.4.1)

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where φ satisfies φ(x0, y0) = 0 and lim(x,y)→(x0,y0) φ(x, y) = 0. Note that∣∣∣∣(x− x0

y − y0

)∣∣∣∣ = |z − z0|.

The linear function is represented by the Jacobian matrix.

Jf (x0, y0) =

(∂xu ∂yu∂xv ∂yv

)(x0, y0).

The partial derivatives of f are denoted by

∂xf =

(∂xu∂xv

), ∂yf =

(∂yu∂yv

). (1.4.2)

Treated as a complex function,

∂xf = ∂xu+ i∂xv, ∂yf = ∂yu+ i∂yv.

Definition 1.4.1 A function f : U → C is complex differentiable at z0 if there exists acomplex number f ′(z0) and a function ψ with ψ(z0) = 0 and limz→z0 ψ(z) = 0, suchthat

f(z) = f(z0) + (df)z0(z − z0) + ψ(z) |z − z0| . (1.4.3)

The number f ′(z0) is the derivative of f at z0.

Equivalently, f is complex differentiable at z0 with derivative f ′(z0) if and only if

f ′(z0) = limw→0

f(w + z0)− f(z0)

w.

Example 1.4.1 f(z) = z is differentiable. So are any polynomials in z.

This follows fromf(z0 + z)− f(z0)

z= 1

and chain rules.

Example 1.4.2 The function f(z) = z is not complex differentiable.

Proof Note

f(z0 + z)− f(z0)

z=z

z=

{1, if Im(z) = 0−1, if Re(z) = 0,

which means limz→0f(z0+z)−f(z0)

z does not exist. �

Definition 1.4.2 A function is differentiable in U if it is differentiable everyewherein U .

Notation. A function f : Rn → Rm is Cr if it is r times differentiable and itspartial derivatives of order less or equal to r are continuous.

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Figure 1.3: Handwriting by Riemann

Theorem 1.4.3 1. If f : U → R is complex differentiable at z0 = x0 + iy0 then fis real differentiable at (x0, y0) and the Cauchy-Riemann Equations hold at z0:

∂xu = ∂yv, ∂yu = −∂xv. (1.4.4)

Also,

f ′(z0) = ∂xu+ i∂xv =1

i(∂yu+ i∂yv).

2. If f : U → C is real differentiable and satisfies the Cauchy-Riemann equationat a point (x0, y0) ∈ U then f is complex differentiable at z0 = x0 + iy0. Inparticular, if u, v : R2 → R2 are C1 functions satisfying the Cauchy-Riemannequations in U then f = u+ iv is complex differentiable in U .

Proof (1) Write f ′(z0) = s+ it. Then by the definition, (1.4.3),

f(z) = f(z0) +

(s −tt s

)(z − z0) + ψ(z) |z − z0| .

This is (1.4.1) with

(df)(x0,y0) =

(s −tt s

).

So f is real differentiable with(∂xu ∂yu∂xv ∂yv

)=

(s −tt s

).

Thus the Cauchy-Riemann equation follows and

f ′(z0) = s+ it = ∂xu+ i∂xv = ∂yv − i∂yu.

(2) We have (1.4.1),

f(x, y) = f(x0, y0) + (df)(x0,y0)

(x− x0

y − y0

)+ φ(x, y)

∣∣∣∣(x− x0

y − y0

)∣∣∣∣ .By the Cauchy-Riemann equation the Jacobian matrix is the following form

J =

(∂xu −∂xv∂xv ∂xu

),

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and represent the complex linear map: multiplication by f ′(z0) := ∂xu+ i∂xv, Hence

f(z) = f(z0) + f ′(z0)(z − z0) + φ(z) |z − z0| .

This implies f is complex differentiable at z0. If u, v are C1, then f = (u, v) isdifferentiable and the previous statement applies. �

The Cauchy Riemann equation can also be written as ∂xf = 1i ∂yf .

1.5 The ∂ and ∂ operatorGiven a function f , we have

f(x, y) = f

(z + z

2,z − z

2i

).

This inspires the notation :

∂z =1

2(∂x +

1

i∂y), ∂z =

1

2(∂x −

1

i∂y). (1.5.1)

It is common to denote ∂z by ∂ and ∂z by ∂. It is clear that ∂zf = 0 is the Cauchy-Riemann equation

We can reformulate the earlier theorem using these notations. Suppose that f iscomplex differentiable at z then f is real differentiable at z and,

∂f(z) = 0, f ′(z) = ∂zf(z).

1.6 Harmonic Functions

Definition 1.6.1 A real valued function u : R2 → R is a harmonic function if ∆u = 0where ∆ = ∂xx + ∂yy is the Laplacian.

Proposition 1.6.1 If u, v areC2 functions and satisfies the Cauchy-Riemann equations

∂xu = ∂yv, ∂yu = −∂xv,

then u, v are harmonic functions. Consequently u, v are C∞.

Proof We differentiate the Cauchy-Riemann equation to see

∂xxu = ∂xyv, ∂yyu = −∂yxv.

Consequently ∂xxu + ∂yyu = 0. Similarly, ∂xxv + ∂yyv = 0. From standard theoryin PDE, a solution of the elliptic equation ∆u = 0 is C∞. �

Later we see that if f is differentiable in a region, it has derivatives of all orders.So the conditions u, v ∈ C2 can be reduced to C1.

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1.7 Holomorphic Functions

Definition 1.7.1 A function f : U → C is said to be holomorphic on U if it is dif-ferentiable at every point of U ; it is holomorphic at z0 if it is holomorphic in a disccontaining z0. A function f : C→ C is said to be entire if it is complex differentiableat every point of C.

For the next corollary we use the following, thia is also Corollary 4.3.2

Proposition 1.7.1 A holomorphic function in a region with vanishing derivative mustbe a constant.

Proof To see this, we first note that f has vanishing Jacobian matrix, and so it deriva-tives along the coordinate directions vanishes. So f is constant along any line segmentparallel with x and y axis. But any two points in a region can be connected by piece-wise line segments parallel to either x and y axis, and so the values of f at these twopoints must be the same. �

Example 1.7.2 Let U be a region in C. If f : U → R is a real valued function, then fis not holomorphic in U unless f is a constant.

Proof Let f = u + iv where v vanishes identically. If f is holomorphic, by theCauchy-Riemann equation, ∂xu = ∂yu = 0 and f ′(z) = ∂yv + i∂xv = 0, and f mustbe a constant. �

1.8 Rules of DifferentiationTheorem 1.8.1 If f, g are differentiable at z0, the derivatives indicated below exist atz0 and the relations stated below hold when evaluated at z0.

1. (kf)′ = kf ′, for any k ∈ C

2. (f + g)′ = f ′ + g′

3. (fg)′ = fg′ + f ′g

4. (f/g)′ = gf ′−fg′g2 provided g(z0) 6= 0.

Theorem 1.8.2 Suppose that g is differentiable at z0 and f is differentiable at g(z0)then the composition f ◦ g is differentiable at z0 and

(f ◦ g)′(z0) = f ′(g(z0)) g′(z0).

Observe that if the Jacobian matrix of a complex differentiable function f repre-sents complex multiplication, i.e. it is of the form

J =

(∂xu ∂yu−∂yu ∂xu

),

14

then so is its inverse:

J−1 =1

det J

(∂xu −∂yu∂yu ∂xu

).

Since f ′(z0) = ∂xu(z0)− i∂yu(z0), and

1

f ′(z0)=

∂xu(z0) + i∂yu(z0)

(∂xu)2(z0) + (∂yu(z0))2=

1

det J(z0)(∂xu(z0) + i∂yu(z0)).

In conclusion, J(z0) represents f ′(z0), J−1(z0) represents 1f ′(z0) .

This leads to the following theorem.

Theorem 1.8.3 Suppose that f : U → C is complex differentiable and u, v havecontinuous partial derivatives. Suppose f ′(z0) 6= 0 for some z0 ∈ U .

• Then there exists a disc U around z0 such that f : U → f(U) is a bijection,f(U) is open and f−1 : f(U)→ U is continuous.

• Furthermore f−1 is complex differentiable on f(U) and

(f−1)′(f(z)) =1

f ′(z), z ∈ U.

Proof The first part of the statement follows from real analysis. To see f−1 is differ-entiable, write w0 = f(z0), w = f(z). Since f and f−1 are continuous, w → w0 isequivalent to z → z0. Since f ′(z0) 6= 0,

f−1(w)− f−1(w0)

w − w0=

1f(z)−f(z0)

z−z0

.

Take w → w0 we see that the limit on the left hand side exists and

f−1(w0) =1

f ′(z0)=

1

f ′(f−1(w)).

�

Remark 1.8.4 Later we see that if f is complex differentiable, it is infinitely differen-tiable. If f is one to one then f ′ does not vanish. See section 8.3.

15

Chapter 2

The Riemann Sphere andMobius Transforms

2.1 Conformal MappingsDefinition 2.1.1 A parameterized curve in the complex plane is a function z : [a, b]→C where [a, b] is closed interval of R.

If z(t) = x(t) + iy(t) its derivative is z′(t) = x′(t) + iy′(t).

Definition 2.1.2 The parameterized curve : [a, b] → C is smooth if z′(t) exists and iscontinuous on [a, b]. We assume furthermore z′(t) 6= 0.

The derivatives at the ends are understood to be one sided derivatives. From now on bya curve we mean a smooth curve.

If z′(t) does not vanish the curve has a tangent at this point, whose direction isdetermined by arg(z′(t)).

Definition 2.1.3 Let z1 : [a1, b1] → C and z2 : [a2, b2] → C be two smooth curvesintersecting at z0. The angle of the two curves is the angle of their derivatives at thispoint.

They are given by the difference of the arguments of their derivatives. If z1(t1) =z2(t2) = z0, their angle at the point z0 is: arg(z′2(t2))− arg(z′1(t1)).

Definition 2.1.4 A map f : U → C is conformal at z0 if it preserves angles, i.e. if z1

and z2 are two curves meeting at z0, the angle from f ◦ z1 to f ◦ z2 at f(z0) are thesame as the angle from z1 to z2 at z0.

Example 2.1.1 The linear map f(z) = kz where k 6= 0, is a conformal map as it iscomposed of scaling by |k| and rotating by the angle arg(k). c.f.Example 1.2.2

Example 2.1.2 f(z) = z is not a conformal map. This map reverses orientation.

Theorem 2.1.3 If f : U → C is holomorphic at z0 and f ′(z0) 6= 0, then f is conformalat z0.

16

Proof Let z1 : [a1, b1] → C and z2 : [a2, b2] → C be two smooth curves intersectingat z0: for t1 ∈ [a1, b1] and t2 ∈ [a2, b2], z1(t1) = z2(t2) = z0. Let γ1(t) = f ◦ z1(t)and γ2(t) = f ◦ z2(t). Since f ′(z0) does not vanish, γ1 and γ2 have well definedtangents which are:

γ′(t1) =d

dt|t=t1f ◦ z1 = f ′(z1(t1))z′1(t1) = f ′(z0)z′1(t1)

γ′(t2) =d

dt|t=t2f ◦ z2 = f ′(z2(t2))z′2(t2) = f ′(z0)z′2(t2).

Multiply z′1(t1) and z′1(t1) by the non-zero complex number f ′(z0) preserves anglesbetween the two vectors, as well as their orientation, c.f. Example 2.1.1, so the anglefrom γ1 to γ2 at f(z0) is the same as the angle from z1 to z2 at z0. �

Note also,arg(γ′(t1)) = arg(f ′(z0)) + arg(z′1(t1)).

2.2 Mobius Transforms (Lecture 5)A polynomial P (z) = a0 + a1z + . . . anz

n, where ai ∈ C, is an entire function. Theroots zn are the zero’s of P . If there are exactly r roots coincide, this root is said tohave order r. In light of Theorem 2.1.3 it is interesting to know where lie the zero’s ofP ′(z).

By the fundamental theorem of Algebra, which we prove later (Theorem 5.7.2),P (z) = 0 has a complete factorisation:

P (z) = an(z − z1) . . . (z − zn).

Suppose that P and Q are two polynomials without common factors and define therational function

f(z) =P (z)

Q(z).

Then f is defined and is complex differentiable everywhere except at the zeros’ of Q.The zero’s of Q are the poles of f . We now look at rational functions with one poleand one zero.

Definition 2.2.1 The following collection of maps are Mobius transforms{az + b

cz + d: ad− bc 6= 0, a, b, c, d,∈ C

}.

If ad− bc = 0, az+bcz+d is a constant function, and are hence excluded. If f(z) = az+bcz+d

is a Mobius transform, its maximal domain is: C \ {−dc}. Since

f ′(z) =ad− bc

(cz + d)26= 0,

f is a conformal map.

17

2.2.1 The Extended Complex PlaneTo make the statements neat we add a point at infinity to C and define the extendedcomplex plane to be

C∗ = C ∪ {∞}

with the convention:

1

0=∞, 1

∞= 0, a+∞ =∞, a−∞ =∞,

and for a 6= 0, a · ∞ =∞ · a =∞.

2.2.2 Properties of Mobius TransformsLet f(z) = az+b

cz+d . We extend the Mobius transform f from C to C∗ by defining:

f(−dc

) =∞, f(∞) =a

cif c 6= 0.

If c = 0, f(z) = 1f (az + b) is defined on the whole plane, then we define

f(∞) =∞, if c = 0.

The function f has an inverse

f−1(w) =dw − b−cw + a

.

Note that multiply a, b, c, d, by a non-zero number λ does not change the function

f(z) =az + b

cz + d=azλ+ bλ

cλz + dλ.

Hence we may eliminate one parameter and assume that ad− bc = 1. We define

M =

{az + b

cz + d: ad− bc = 1, a, b, c, d,∈ C

}.

Theorem 2.2.1 The setM of Mobius transforms is a group under composition. EachMobius transform is a composition of the following maps:

(1) translation: z 7→ z + a for some complex number a;

(2) composition of scaling and rotation:

z 7→ kz, some k ∈ C, k 6= 0.

(3) Inversion: z 7→ 1z .

Proof For the group we check the following:

• f(z) = z is the identity. (a = 1, b = c = 0, d = 1)

• If f(z) = az+bcz+d ∈M, then

f−1(w) =dw − b−cw + a

∈M.

18

• If f(z) = az+bcz+d ∈M and g(z) = az+b

cz+d∈M. Then f ◦ g = Az+B

Cz+D ∈M wherethe complex numbers A,B,C,D are given by(

A BC D

)=

(a bc d

)(a bc d

).

For the second part of the statement, if c = 0, az+bd = adz + b

d . If c 6= 0,

f(z) =az + b

cz + d=a

c+bc− adc2

1

z + dc

.

�

Example 2.2.2 The map f(z) = z+1z−1 is called the Cayley transform. It takes C \ {1}

to itself, f : C \ {1} → C \ {1} is a bijection and f−1 = f . Let us consider f as a mapon C∗ by setting f(1) =∞, f(∞) = 1. Note

f(x+ iy) =x2 + y2 − 1

(x− 1)2 + y2+

−2y

(x− 1)2 + y2i.

Let γ = {x2 + y2 = 1} with γ+ and γ− denote respectively the upper and lower halfof the circle. Then,

• f sends {−1, 0, 1} to {0,−1,∞} respectively.

• f sends the upper circle to the lower half of the imaginary axis.

• f sends x-axis to x-axis. It send the x-axis within the unit disc to the negativex-axis.

• f sends the upper half of the unit disc to the third quadrant.

• f sends the lower circle to the upper half of the imaginary axis.

• f sends the lower half of the unit disc to the second quadrant.

• f sends the exterior of the unit circle to the right half of the plane.

If z2, z3, z4 are distinctive points in C∗ we associate to it the Mobius transform

f(z) =z − z3

z − z4/z2 − z3

z2 − z4=

(z − z3)(z2 − z4)

(z − z4)(z2 − z3), if z2, z3, z4 ∈ C.

If one of these points is the point at infinity the map is interpreted as following:

f(z) =

z − z3

z − z4, if z2 =∞

z2 − z4

z − z4, if z3 =∞

z − z3

z2 − z3, if z4 =∞.

Note that if z2, z3, z4 ∈ C,

f(z2) = 1, f(z3) = 0, f(z4) =∞.

19

Also,If z2 =∞, f(∞) = 1, f(z3) = 0, f(z4) =∞,If z3 =∞, f(z2) = 1, f(∞) = 0, f(z4) =∞,If z4 =∞, f(z2) = 1, f(z3) = 0, f(∞) =∞.

We denote by Fz2z3z4 this map. Note it sends {z2, z3, z4} to {1, 0,∞}.

Lemma 2.2.3 A Mobius transform can have at most two fixed points unless f(z) isthe identity map.

Proof We solve for az+bcz+d = z, equivalently cz2 + (d− a)z − b = 0. This has at mosttwo solutions(use polynomial long division/ the Euclidean algorithm). �

Proposition 2.2.4 For any two sets of distinctive complex numbers {z2, z3, z4} and{w2, w3, w4} in C∗, there exists a unique Mobius transform taking zi to wi for i =2, 3, 4.

Proof We knowFz2z3z4 takes {z2, z3, z4} to {1, 0,∞}, and the inverse map ofFw2w3w4

takes {1, 0,∞} to {w2, w3, w4}. The composition F−1w2w3w4

◦Fz2z3z4 takes {z2, z3, z4}to {w2, w3, w4}.

To prove this map is unique, suppose f, g are two Mobiums transform sending{z2, z3, z4} to {w2, w3, w4}. Then f ◦ g−1(wi) = f(zi) = wi. The Mobius transformf ◦ g−1 has three fixed points: w1, w2, w3. By Lemma 2.2.3, f ◦ g−1 is the identitymap and f = g identically. �

Corollary 2.2.5 For any distinctive complex numbers {z2, z3, z4} in C∗, the Mobiustransform Fz2,z3,z4 is the only Mobius transform that takes {z2, z3, z4} to {1, 0,∞}.

Proposition 2.2.6 Let r, c be real numbers, k ∈ C. Then the equation

r|z|2 + kz + kz + c = 0

• represents a line if r = 0 and k 6= 0.

• a circle if r 6= 0, and |k|2 ≥ rc.

• The circle equation is |z + kr | = 1

r

√|k|2 − rc, whose locus is an emptyset if

r 6= 0 and |k|2 < rc.

This is clear by expanding z = x+ iy in x and y.

Definition 2.2.2 The locus of the points of r|z|2 − kz − kz + c = 0, if non-empty, iscalled a circleline.

We see later this definition is not merely a simplification of terminologies. Both circlesand extended lines in the plane correspond to circles in the Riemann sphere.

Lemma 2.2.7 Let R be a real number, z1, z2 complex numbers. The locus of theequation

|z − z1| = r|z − z2|

20

represents a circle if r 6= 1. If z = x+ iy, z1 = x1 + iy1 and z2 = x2 + iy2 then

(x− x1)2 + (y − y1)2 = r(x− x2)2 + r(y − y2)2.

If r = 1, the equation is

2(x2 − x1)x+ 2(y2 − y1)y = |z2|2 − |z1|2,

representing a line if z1 6= z2. It is the set of points which are equi-distance from z1

and z2, i.e. the line perpendicular to the line segment [a, b] and passing its mid-point.

Proposition 2.2.8 A Mobius transform maps a circleline to a circleline.

Proof Since a Mobius transform is the composition of translation, multiplication bya non-zero complex number and inversion we only need to prove it for each of thesemaps. The inverse of such transformations are of the same type.

A translation, z 7→ z + a, takes a circleline to a circleline: the image of

r|z|2 + kz + kz + C = 0,

in the z-plane is precisely the locus of the equation below in the w-plane:

r|w − a|2 + k(w − a) + k(w − a) + C = 0,

i.e.r|w|2 + (k − a)w + (k − a)w + r|a|2 − (ka+ ka) + C = 0.

Note that r|a|2− (ka+ka) +C is a real number. Multiplication by a complex numberis a composition of scaling with rotation, it clearly takes a circleline to a circleline.Finally we work with the inversion z 7→ 1

z . It takes |z| = r to |z| = 1r trivially. Let

a 6= 0. If w is in the image of |z−a| = r then | 1w −a| = r, i.e. |w− 1a | =

r|a| |w| which

is a circleline, by Lemma 2.2.7. Let us take a line kz+kz+C = 0. The equation of itsimage w = 1

z satisfies k 1w +k 1

w +C = 0 which is equivalent to kw+kw+C|w|2 = 0representing a circle if C 6= 0 and a line otherwise. �

Exercise 2.2.9 Given r, c ∈ R and k ∈ C, and the equation

r|z|2 − kz − kz + c = 0.

Identify its image under the transformw = k′z where k′ is a non-zero complex number.

Definition 2.2.3 The cross ratio of z1, z2, z3, z4, denoted by [z1, z2, z3, z4], is the com-plex number:

[z1, z2, z3, z4] := Fz2z3z4(z1).

In other words, it is

[z1, z2, z3, z4] =(z1 − z3)(z2 − z4)

(z1 − z4)(z2 − z3),

interpreted appropriately if one of them is∞.

Proposition 2.2.10 Let z1, z2, z3, z4 be distinct points in C∗. Then [z1, z2, z3, z4] is areal number if the four points lie in a circle or on the extended line R ∪ {∞}.

21

Proof If [z1, z2, z3, z4] is a real number, Fz2,z3,z4 maps the four points z1, z2, z3, z4 torespectively [z1, z2, z3, z4], 1, 0,∞, all on the extended x-axis. The map (Fz2,z3,z4)−1

takes the 4 points [z1, z2, z3, z4], 1, 0,∞ back to z1, z2, z3, z4. Note a Mobius trans-form takes the extended line to a circleline, so the four points lie in a circleline.

If the four points lie in a circleline, then the map Fz2,z3,z4 takes the circlelineto a circleline. This will be the line determined by (1, 0,∞), the x-axis. HenceFz2,z3,z4(z1) must be a real number.

�

2.3 The Riemann Sphere and Stereographic Projection(lecture 7)

The purpose of the section is to give a concrete geometric representation of the ex-tended plane as the Riemann sphere. In particular we observe that the point at infinityis just represented as a point in the sphere. Let us denote by S2 the unit sphere in R3:

S2 = {(X,Y, Z) : X2 + Y 2 + Z2 = 1}.

We fix the north poleN = (0, 0, 1) and associate with each P on S2\{N}with a pointπ(P ) on the plane which is the intersection of the line from N to P with the plane.

Proposition 2.3.1 The stereographic projection from S2 → C∗ is :

π((X,Y, Z)) =X + iY

1− Z, π(N) =∞. (2.3.1)

The inverse map is given by

π−1(z) = (2Re(z)

|z|2 + 1,

2Im(z)

|z|2 + 1,|z|2 − 1

|z|2 + 1). (2.3.2)

Proof Suppose that P = (X,Y, Z), write (x, y, 0) = π(P ). The line equation con-necting N,P and π(P ) is given by:

(x, y, z) = (0, 0, 1) + t(X − 0, Y − 0, Z − 1).

Setting z = 0 we see

t =1

1− Z, x = tX =

X

1− Z, y = tY =

Y

1− Z, (2.3.3)

22

proving π((X,Y, Z)) = X+iY1−Z . Let z = x + iy be a point in C, we find its inverse

π−1(z). We uset2(X2 + Y 2 + Z2) = t2.

Thus

x2 + y2 = t2(1− Z2) =1− Z2

(1− Z)2=

1 + Z

1− Z.

Finally

Z =x2 + y2 − 1

x2 + y2 + 1=|z|2 − 1

|z|2 + 1.

By t = 11−Z and (2.3.3), we see X = x(1−Z) = 2x

|z|2+1 , Y = y(1−Z) = 2y|z|2+1 . �

The following is an easy exercise.

Proposition 2.3.2 The antipodal point to a point (X,Y, Z) in S2 is (−X,−Y,−Z). Ifz ∈ C∗ corresponds to a point in S3 then − 1

z corresponds to the antipodal point in S2.

Definition 2.3.1 If z1, z2 ∈ C∗ we define the stereographic distance to be

d(z1, z2) = |π−1(z1)− π−1(z2)|.

If p = (X,Y, Z) and p′ = (X ′, Y ′, Z ′) are points in the sphere, their distance is:

|P −P ′| = |X−X ′|2 + |Y −Y ′|2 + |Z−Z ′|2 = 2−2(XX ′+Y Y ′+ZZ ′) (2.3.4)

If z′ =∞, then π−1(z′) = (0, 0, 1), X ′ = 0, Y ′ = 0 and Z ′ = 1. Consequently,

d(z,∞) =

√2− 2

|z|2 − 1

|z|2 + 1=

2√|z|2 + 1

.

This agrees with the intuition, z →∞means |z| → ∞. If z, z′ ∈ C, apply (2.3.4), anduse (2.3.2) we see

(d(z, z′))2 = 2− 2

((z + z)

|z|2 + 1· (z′ + z′)

|z′|2 + 1+

z−zi

|z|2 + 1·

z′−z′i

|z′|2 + 1+|z|2 − 1

|z|2 + 1· |z′|2 − 1

|z′|2 + 1

)

Since

(|z|2 + 1)(|z′|2 + 1)− (|z|2 − 1)(|z′|2 − 1)

= 2|z|2 + 2|z′|2(z + z)(z′ + z′)− (z − z)(z′ − z′) = 2zz′ + 2zz′.

Also, |z|2 + |z′|2 − zz′ − zz′ = (z − z′)(z − z′) = |z − z′|2. Cleaning up the righthand side we obtain:

d(z, z′) =2|z − z′|√

|z|2 + 1√|z′|2 + 1

.

Definition 2.3.2 The space S2 is the Riemann sphere. A circle on S2 is the intersectionof a plane with S2.

Proposition 2.3.3 A circle on S2 corresponds to a circle or a line on C∗.

23

Proof let us take a plane:

AX +BY + CZ +D = 0

where A,B,C,D are real numbers. Note that the north pole passes through the planeif and only C +D = 0.

Let z = x+iy ∈ C. Then a point π−1(z) = ( 2x|z|2+1 ,

2y|z|2+1 ,

|z|2−1|z|2+1 ) on S2 satisfies

the plane equation if and only if

2xA+ 2BY + (|z|2 − 1)C + (|z|2 + 1)D = 0.

Rearrange the equation:

(C +D)(x2 + y2) + 2xA+ 2By + (D − C) = 0, (2.3.5)

which represents a circle in the plane or an empty set when C + D 6= 0. If the planeintersects with S2, it is not empty and so is a circle. If C +D = 0 this is a line on theplane.

Let us consider a circle or an extended line in C∗. It is of the form:

A(x2 + y2) + Bx+ Cy + D = 0 (2.3.6)

where A, B, C, D are real numbers. Let us solve for A,B,C,D:

C +D = A, 2A = B, 2B = C,D − C = D.

Then (2.3.6) is equivalent to (2.3.5) which means the corresponding points of the cir-cleline on the plane satisfies

AX +BY + CZ +D = 0,

and their image by π−1 line on a circle in S2. �

If the plane passes through the origin we have a great circle. This is so if and onlyif D = 0 and we have

(x2 + y2) +2A

Cx+

2B

Cy = 1.

The plane passes through the north pole if and only if C + D = 0 in which casethe circle projects to a line.

24

Chapter 3

Power Series

Definition 3.0.1 A series of complex numbers∑∞n=0 an is said to converge if the par-

tial sum∑Nn=0 an converge. It is said to converge absolutely if

∑∞n=0 |an| converges.

Evidently∑∞n=0 an is convergent is equivalent to both

∑∞n=0 Re(an) and

∑∞n=0 Im(an)

converge.

Proposition 3.0.1 If∑∞n=0 an converges absolutely, then it is convergent.

Just note that |Re(an)| ≤ |an| and |Im(an)| ≤ |an|. Follow this with the standardcomparison test.

3.1 Power series is holomorphic in its disc of conver-gence

Let us consider a power series∑∞n=0 an(z − z0)n where an, z0 and z are complex

numbers. For simplicity let us take z0 = 0.

Theorem 3.1.1 Let∑∞n=0 anz

n be a power series where an ∈ C. There exists R ∈[0,∞], such that the following holds:

(1) If |z| < R, the series converges absolutely.

(2) If |z| > R, the series diverges.

Moreover, there is Hadamard’s formula:

1

R= lim sup

n→∞(|an|)

1n

with the convention 1∞ = 0 and 1

0 = ∞. The region {|z| < R} is called the disc ofconvergence and R its radius of convergence.

Proof Suppose A = lim supn→∞(|an|)1n is such that 0 < A <∞. Then there exists

N such that for n ≥ N , |an|1n ≤ A. If |z| < 1

A , then there exists δ > 0 such that|z| < 1

A+δ and for n ≥ N , |an|1n |z| ≤ A

A+δ < 1, and∑∞n=0 |an||z|n is convergent. If

25

|z| > 1A , there exists 0 < δ < A such that |z| > 1

A−δ and |an|1n |z| ≥ A

A−δ > 1 forn ≥ N , it follows that

∑∞n=0 anz

n is divergent.If lim supn→∞(|an|)

1n = ∞, then for any non-zero z, |an||z|n does not converge

to 0 as n → ∞ and the power series is divergent for all z 6= 0. ( There is a sequenceank with |ank |

1nk > 2

|z| ).

If lim supn→∞(|an|)1n = 0, then for any |z| and for any 0 < ε < 1

2|z| there exists

N such that |an|1n ≤ ε, for all n ≥ N , and

|an||z|n ≤ εn|z|n ≤1

2n.

The power series converges absolutely for any z. �

By composing with translation z → z − z0, we may translate the statement of thetheorem from the power series

∑∞n=0 an|z|n to the power series

∑∞n=0 an|z − z0|n.

Theorem 3.1.2 The power series f(z) =∑∞n=0 an(z − z0)n defines a holomorphic

function in its disc of convergence. Furthermore,

f ′(z) =

∞∑n=1

nan(z − z0)n−1

and f ′ has the same radius of convergence as f .

Proof If R is the radius of convergence for f , then using Hadamard’s formula we seethe radius of convergence for

∑∞n=1 nan(z − z0)n−1 is R. Take z from its disc of

convergence, {z : |z − z0| < R}. Define

fN (z) =

N∑n=0

an(z − z0)n, f ′N (z) =

N−1∑n=1

nan(z − z0)n−1.

Then for h ∈ C,∣∣∣∣∣f(z + h)− f(z)

h−∞∑n=1

nan(z − z0)n−1

∣∣∣∣∣≤∣∣∣∣fN (z + h)− fN (z)

h− f ′N (z)

∣∣∣∣+

∞∑n=N+1

∣∣∣∣an(z + h− z0)n − an(z − z0)n

h

∣∣∣∣+

∞∑n=N

n|an(z − z0)n−1|.

The last term on the right hand side is the remainder term of the convergent series∑∞n=1 nan(z − z0)n−1. Given ε > 0, there exists N0 such that if n ≥ N0, this last

term is less than ε/3. Furthermore there exists a number δ0 > 0 and 0 < A < R suchthat |z + h− z0| < A for |h| ≤ δ0. In the following we use the identity:an − bn = (a− b)(an−1 + an−2b+ . . .+ bn−1),∞∑

n=N+1

∣∣∣∣an(z + h− z0)n − an(z − z0)n

h

∣∣∣∣ ≤ ∞∑n=N+1

|an|∣∣∣∣ (z + h− z0)n − (z − z0)n

h

∣∣∣∣26

≤∞∑

n=N+1

|an|∣∣(z + h− z0)n−1 + (z + h− z0)n−2(z − z0) + . . .+ (z − z0)n−1

∣∣≤

∞∑n=N+1

|an|nAn.

The right hand side is again the tail of a convergent series, hence there exists N1 >N0 such that

∑∞n=N1

|an|nAn−1 ≤ ε/3. Finally, fN1is a differentiable function,

limh→0

∣∣∣∣fN1(z + h)− fN1(z)

h− SN (z)

∣∣∣∣ = 0.

By choosing h sufficiently small,∣∣∣ fN1

(z+h)−fN1(z)

h − SN (z)∣∣∣ < 1

3ε. The proof iscomplete. �

Corollary 3.1.3 A power series function f(z) =∑∞n=0 an(z − z0)n is infinitely dif-

ferentiable in its disc of convergence. Furthermore an = f(n)(z0)n! .

3.2 Analytic Continuation (Lecture 8)Definition 3.2.1 A function f : U → C is said to be analytic, or has a power seriesexpansion, at z0 ∈ U , if there exists a power series with positive radius of convergencesuch that

f(z) =

∞∑n=0

an(z − z0)n, for z in a neighbourhood of z0.

We say f is analytic on U if it has a power series expansion at every point of U .

Example 3.2.1 The function f(z) = 1z is analytic on C \ {0}. Note 1

1−z =∑∞n=0 z

n

for all |z| < 1. It is clear f has the power series expansion at z = 1:

∞∑n=0

(1− z)n, |z − 1| < 1.

If z0 is any non-zero number, take z with |z0 − z| < |z0|, then

1

z=

1

z0 − (z0 − z)=

1

z0· 1

1− z0−zz0

=

∞∑n=0

(−1)n(z0)−n−1(z − z0)n.

The power series converges for any z with |z0 − z| < |z0| (in particular z 6= 0). Hencef is analytic.

Theorem 3.2.2 The power series function f(z) =∑∞n=0 an(z−z0)n is analytic in its

disc of convergence D = {z : |z − z0| < R}. In fact for w ∈ D,

f(z) =

∞∑n=0

f (n)(w)

n!(z − w)n, ∀z ∈ D(w,R− |w − z0|).

27

Proof Take w ∈ D. Note that |w− z0| < R and let z satisfy |z −w| < R− |w− z0|.We expand the power series

f(z) =

∞∑n=0

an(z − w + w − z0)n =

∞∑n=0

(an

n∑k=0

(nk

)(z − w)k(w − z0)n−k

)

=

∞∑k=0

( ∞∑n=k

(nk

)an(w − z0)n−k

)(z − w)k.

To justify the exchange of the order in the above computation, we bound the the partialsum and use the rearrangement of double series lemma below:

N∑n=0

n∑k=0

|an|(nk

)|z − w|k|w − z0|n−k =

N∑n=0

|an|(|z − w|+ |w − z0|)n

≤∞∑n=0

|an|(|z − w|+ |w − z0|)n <∞∑n=0

|an|rN <∞,

where r is a number smaller than R. Hence the partial sum up to N is bounded by∑∞n=0 |an|rN . Then

f(z) =

∞∑k=0

bk(z − w)k, where bk =

∞∑n=k

an

(nk

)(w − z0)n−k.

It is clear f (k)(w) = bkk!. �

Lemma 3.2.3 (Double Series Lemma) Suppose there exists a number M such that

N∑i=0

N∑j=0

|aij | ≤M

Then all linear arrangements of the double series converge absolutely to the same num-ber.

a00 + a01 + a02 + a03 + . . . = s0

+ + + + +a10 + a11 + a12 + a13 + . . . = s1

+ + + + +a20 + a21 + a22 + a23 + . . . = s2

+ + + + +a30 + a31 + a32 + a33 + . . . = s3

+ + + + +: : : : := = = = =v0 + v1 + v2 + v3 + . . . = ???

(3.2.1)

Figure 3.1: Figure From E. Hairer and G. Wanner

28

Definition 3.2.2 If f : V → C is a function where V is a subset of C and g : U → Cis an analytic function in a region U with V ⊂ U . If f, g agree on V we say g is ananalytic continuation of f into the region U .

The set V is not required to be a region. We may wonder which functions has ananalytic continuation.

Example 3.2.4 If f(x) =∑an(x − x0)n is a real power series function with radius

of convergence R. We define g(z) =∑an(z − x0)n. Then g has the same radius of

convergence R ( use Hadamard’s formula for R). So g is an analytic continuation (alsoknown as analytic extension) of f from (−R,R) to the disc {z : |z| < R}.

Example 3.2.5 If P (x) is a polynomial with one real variable then P (z) is its analyticcontinuation into C.

Example 3.2.6 Let h(z) =∑∞n=0(1− z)n, |z− 1| < 1. Then f(z) = 1

z is an analyticcontinuation of h into the punctured complex plane C \ {0}.

Later we will study zeros of analytic function and conclude a function definedon any connected set containing an accumulation point can have only one analyticcontinuation.

3.3 The Exponential and Trigonometric FunctionsThe exponential functions and trigonometric functions are analytic continuations oftheir corresponding functions on the real line.

Definition 3.3.1 We define the following function by power series:

ez =

∞∑n=0

zn

n!, sin(z) =

∞∑n=0

(−1)nz2n+1

(2n+ 1)!, cos(z) =

∞∑n=0

(−1)nz2n

(2n)!.

They are entire functions. By adding two series together we see that

sin(z) =eiz − e−iz

2i, cos(z) =

eiz + e−iz

2,

and Euler’s formula:

eiz = cos(z) + i sin(z).

We also define the following functions: sinh(z) =∑∞n=0

z2n+1

(2n+1)! and cosh(z) =∑∞n=0

z2n

(2n)! . Note that sinh(z) = ez−e−z2 and cosh(z) = ez+e−z

2 . Most propertiesfor the corresponding real trigonometric functions are inherited by the complex valuedtrigonometric functions. For example the zeros of sin(z) are at nπ. But sin(z) is not abounded function, nor is cos(z).

29

Theorem 3.3.1 A power series f(z) =∑∞n=0

zn

n! satisfies

f(z + w) = f(z)f(w)

for all z, w, z + w. In particular,

e2kπi = 1, ez+2kπi = ez, k = 0,±1,±2, . . . .

Note that ex+iy = exeiy and eiy for y ∈ (−π, π) traces out a circle without thepoint on the left real axis. Let

U = {z : −π < Imz < π}

Thenez : U → C \ {re±iπ : r ≥ 0}

is a bijection.

Figure 3.2: Graph by E. Hairer and G. Wanner

3.4 The Logarithmic Function and Power Function (Lec-ture 9)

Definition 3.4.1 The principal branch of the logarithm is the inverse of ez on the slitplane C \ {re±iπ : r ≥ 0}. log(z) = log |z|+ i arg(z), arg(z) ∈ (−π, π).

Theorem 3.4.1 The logarithmic function defined above is holomorphic on its domainof definition and (log z)′ = 1

z .

Proof Apply Theorem 1.8.2 to the exponential function from U to the slit domain. �

Other branches of log z include: log z = log |z|+i arg(z)+2kπ, arg(z) ∈ (−π, π).

Definition 3.4.2 For λ ∈ C we define zλ = eλ log z .

30

Chapter 4

Complex Integration

If a continuous function has a primitive in a region, then its integral along any closedpiecewise smooth curve vanishes. The converse holds in a star region (more generally,in a simply connected region): if a continuous function integrate to zero along anytriangle inside the region, then it has a primitive. Goursat’s theorem states that the in-tegral of a holomorphic function in a region indeed integrate to zero along any trianglewho and whose interior is contained in the region. From this we see Cauchy’s theoremfor a star region: if f is holomorphic in a star region, then it integrates to zero alongany closed smooth curve γ whose interior is contained entirely in the region. Cauchy’stheorem is in fact valid for any simply connected region.

Every point in a region U has a disc around it, contained entirely in U . So ev-ery point in U has a star region neighbourhood (i.e. a region is ‘locally simply con-nected’). In another word, a holomorphic function integrates to zero along any closedsmooth curve with sufficiently small enclosure. The distinction between the local andthe global null integral property relates to homotopy theory as well as to de Rham’scohomology theory built on closed and exact differential forms, both are related to theconcept of ‘simply connectedness of a region’. On the complex plane, the simply con-nectedness can be explained visually which is also responsible for the beauty of thetheory on the plane. It is perhaps confusing in the beginning when confronted withdifferent version’s and various forms of Cauchy’s formulas, the rule of thumb is thefollowing. That a function is differentiable at a point is a local property, we could pickour disc as small as we like. A value for an integral along a closed interval is a globalproperty: it depends on the region enclosed by the curve.

4.1 Complex IntegrationBy a C1 function we mean a differentiable function with continuous derivative. Thederivatives at the ends of an interval are one sided derivatives.

Definition 4.1.1 A parameterized smooth curve is a C1 function, z : [a, b] → C. It ispiecewise smooth if z is continuous on [a, b] and there exist ti s.t. a = t0 < t1 < · · · <tn = b s.t. z is smooth on each sub-interval [ti, ti+1].

A piecewise smooth curve consisting of a finite number of smooth pieces, joined at theends. We sometimes abbreviate ‘a piecewise smooth curve’ to ‘a smooth curve’. A

31

parameterized curve has an orientation: it is the direction a point on the curve travelsas the parameter t increases.

Definition 4.1.2 Two parameterizations γ : [a, b] → C and γ′ : [a′, b′] → C areequivalent if there exists a C1 bijection α : [a, b] → [a′, b′] such that α′(t) > 0 andγ = γ′ ◦ α.

The condition α′(t) > 0 means orientation is preserved. The family of all equivalentparameterizations determine a smooth oriented curve.

Example 4.1.1 The circle {|z − z0| = r} has the obvious parameterizations:

z = z0 + eiθ, 0 ≤ θ ≤ 2π.

The orientation of the curve is anticlockwise (it is a positively oriented curve). Thecurve

z = z0 + e−iθ, 0 ≤ θ ≤ 2π

is negatively oriented.

If f : [a, b]→ C is a continuous function and f(t) = u(t) + iv(t) then∫ b

a

f(t)dt =

∫ b

a

u(t)dt+ i

∫ b

a

v(t)dt.

Lemma 4.1.2 ∣∣∣∣∣∫ b

a

f(t)dt

∣∣∣∣∣ ≤∫ b

a

|f(t)|dt.

Proof Let θ be the principle argument of the complex number∫ baf(t)dt. Then∣∣∣∣∣

∫ b

a

f(t)dt

∣∣∣∣∣ = e−iθ∫ b

a

f(t)dt =

∫ b

a

e−iθf(t)dt = Re

(∫ b

a

e−iθf(t)dt

)

=

∫ b

a

Re(e−iθf(t)

)dt ≤

∫ b

a

|f(t)|dt.

�

4.2 Integration Along a Curve (lecture 9)

Definition 4.2.1 Let f : U → C be a continuous function. Let γ be a smooth curvecontained in U with parameterization z : [a, b]→ C. We define the integral of f alongγ to be: ∫

γ

f(z)dz =

∫ b

a

f(z(t))z(t)dt.

32

Let us write f = u+ iv and z(t) = x(t) + iy(t). Then

f(z(t))z(t) = u(z(t))x(t)− v(z(t))y(t) + i (u(z(t))y(t) + v(z(t))x(t)) .

Hence ∫γ

f(z)dz =

∫ b

a

(u(x(t), y(t))x(t)− v(x(t), y(t))y(t)

)dt+

i

∫ b

a

(u(x(t), y(t))y(t) + v(x(t), y(t))x(t)

)dt.

The integral∫γf(z)dz can also be defined directly by the Riemann sums:

n−1∑i=0

f(z(si))(z(si+1)− z(si))

where a = s0 < s1 < · · · < sn = b is a partition on [a, b]. If the Riemann sumconverges as the size of the partition converges to zero, the limit is

∫ baf(z)dz.

Proposition 4.2.1 The integral∫γf(z)dz is independent of the parameterization.

Proof Let z : [a, b] → C and z : [a′, b′] → C be two parameterizations of γ. Letα : [a′, b′]→ [a, b] be a C1 bijection such that α′(t) > 0 and z = z ◦ α. Then∫ b′

a′f(z(t))

d

dtz(t)dt =

∫ b′

a′f(z ◦ α(t))

d

dtz(α(t))dt

=

∫ b′

a′f(z ◦ α(t))z(α(t))α(t)dt =

∫ b

a

f(z(s))z(s)ds.

�

Definition 4.2.2 The length of the curve γ is:

length(γ) =

∫ b

a

|z′(t)|dt.

The length of the curve is also independent of the parameterization, be an argumentsimilar to that in the proposition above.

Definition 4.2.3 On a piecewise smooth curve γ, consisting of a finite number ofsmooth curves γi, we define∫

γ

f(z)dz =∑i

∫γi

f(z)dz.

Theorem 4.2.2 The following properties hold.(1) ∫

γ

(k1f + k2g)(z)dz = k1

∫γ

f(z)dz + k2

∫γ

g(z)dz.

33

(2) ∫γ

f(z)dz = −∫γ−f(z)dz,

where γ− is the curve γ with reversed orientation, e.g. z−(t) = z(a+ b− t).(3) ∣∣∣∣∫

γ

f(z)dz

∣∣∣∣ ≤ supz∈γ|f(z)| · length of (γ).

A curve is simple if it does not intersect with itself except at the end points. Wewould be interested in the contour of a region, e.g. the contour of a disc is the cir-cle, traveled anticlockwise once. By a circle, we usually mean traveling along it, anticlockwise, once.

Example 4.2.3 Let γ be the circle |z − z0| = r with positive orientation. Let n be aninteger. Then ∫

γ

(z − z0)ndz =

{0, n 6= −12πi, n = −1.

Proof Let us take the parameterization z = z0 + reit, 0 ≤ t ≤ 2π.∫γ

(z − z0)ndz =

∫ 2π

0

(reit)nd

dt(reit)dt

= rn+1i

∫ 2π

0

ei(n+1)tdt

=

{rn+1i

∫ 2π

0cos((n+ 1)t)dt− rn+1

∫ 2π

0sin((n+ 1)t)dt = 0, n 6= −1

i∫ 2π

0dt = 2πi, n = −1.

�

4.3 Existence of Primitives (Lecture 9)

Definition 4.3.1 A curve γ : [a, b]→ C is closed if γ(a) = γ(b).

Definition 4.3.2 A function f : U → C where U is a region is said to have a primitiveF if F : U → C is holomorphic on U and F ′ = f .

The Fundamental theorem of Calculus states that if g : [a, b] → R is a continuousfunctions, then G′(x) = g(x) where G(x) =

∫ xag(t)dt. Unlike for real differentiabil-

ity, the existence of a primitive is a much stronger property. To begin with, given twopoints on a plane, there are many paths leading from one to the other. This is relatedto a path independent property. Later we see that if f has a primitive in a region, it isitself complex differentiable in this region.

Rule of thumb. If you have a real valued function g of one variable x with explicitformulation which does not involve the number i, denote f the function obtained byreplacing x with z. Then f is a ‘candidate analytic continuation’ of g. Compute theformal anti-derivative of g and denote it by G. Now in G replace x by z and denote

34

the function by F . Then F is a ‘candidate primitive’ for f . Once you have a primitive,check it out: is it actually complex differentiable in the desired region? If so, computeits derivative by the rule of calculus.

Theorem 4.3.1 If f : U → C is continuous and has a primitive F , if γ is a piecewiseC1 curve in U begins at w1 and ends at w2 then∫

γ

f(z)dz = F (w2)− F (w1).

In particular, if γ is a closed curve then∫γ

f(z)dz = 0.

Proof Let z : [a, b]→ C be a parameterization of γ. We first assume that it is C1 anddiscuss the piecewise C1 curve later. Then∫

γ

f(z)dz =

∫ b

a

f(z(t))z′(t)dt =

∫ b

a

F ′(z(t))z′(t)dt

=

∫ b

a

d

dtF (z(t))dt = F (z(b))− F (z(a)) = F (w2)− F (w1).

If γ is a piecewise smooth curve, joined by smooth curves on each subinterval of t0 =a < t1 < · · · < tn = b, then we have a telescopic sum:∫

γ

f(z)dz

=[F (z(tn))− F (z(tn−1))] + · · ·+ [F (z(t2))− F (z(t1))] + [F (z(t1))− F (z(t0))]

=F (z(b))− F (z(a)) = F (w2)− F (w1).

�

Corollary 4.3.2 If f : U → C is holomorphic, where U is a connected open set, andf ′ = 0 identically on U then f is a constant.

Proof Let us fix a point z0 ∈ U . Let z ∈ U be any other point. Recall U is pathconnected: we can connect z0 to z by a piecewise smooth curve γ. Since f ′ = 0 is acontinuous function,

f(z)− f(z0) =

∫γ

f ′(z)dz = 0.

�

The converse to the vanishing of integral theorem is at the heart of complex inte-gration, for which we must restrict the region. We work with a sub class of regions ofthe simply connected region, called the star regions.

Definition 4.3.3 A region U is said to be a star region if it has a centre C, by whichwe mean for all z ∈ U , the line segment

{(1− t)C + tz : 0 ≤ t ≤ 1}

belongs to U .

35

Note that z(t) = (1− t)C + tz is a prameterization of the line segment from C to z.Star regions include discs, triangles, rectangles and more generally convex sets. A

star region is simply connected, by which we mean any simple closed curve in thatregion can be continuously deformed to a point. Many regions such as polygons canbe divided into star regions, which allow us to conclude statements for star regions formore general regions.

Figure 4.1: Graph by E. Hairer and G. Wanner

Let us denote by [C, z] the line segment from C the z.

Theorem 4.3.3 (Integrability Criterion) Let U be a star region with a centre C andf : U → C a continuous function. Suppose that∫

T

f(ζ)dζ = 0

for any triangle T contained entirely in U , and with C one of its vertex. Then f has aprimitive in U . In particular

F (z) =

∫[c,z]

f(ζ)dζ, z ∈ U.

Proof The line segment [c, z] is contained in U , so F (z) =∫

[c,z]f(ζ)dζ makes

sense. Let z0 ∈ U . Take z close to z0 so that the triangle region with vertex z0,z and C are contained entirely in U . Denote T this triangle. By the assumption,

36

0 =

∫T

f(ζ)dζ =

∫[z0,C]

f(ζ)dζ +

∫[C,z]

f(ζ)dζ +

∫[z,z0]

f(ζ)dζ.

ThusF (z) = F (z0)−

∫[z,z0]

f(ζ)dζ.

Since −∫

[z,z0]f(z0)dζ =

∫[z0,z]

f(z0)dζ = f(z0)(z − z0),

F (z) = F (z0) + f(z0)(z − z0) +

∫[z0,z]

(f(ζ)− f(z0))dζ.

Define ψ(z0) = 0 and for z 6= z0,

ψ(z) =

∫[z0,z]

(f(ζ)− f(z0))dζ

|z − z0|.

Then

|ψ(z)| =

∣∣∣∣∣∫

[z0,z](f(ζ)− f(z0))dζ

|z − z0|

∣∣∣∣∣ ≤ maxζ∈[z0,z]

|(f(ζ)− f(z0))|.

Since f is continuous, limz→z0 maxζ∈[z0,z] |(f(ζ) − f(z0))| = 0. This proves that Fis differentiable at z0 with F ′(z0) = f(z0). Since z0 is an arbitrary point in U , F isholomorphic in U and a primitive of f . �

4.4 Goursat’s LemmaWe begin with proving a special version of Cauchy’s theorem for triangles whose proofis historical and illuminating. We assume the triangle and its interior is contained in anopen set on which f is holomorphic, hence we do not need to assume U itself is simplyconnected.

Theorem 4.4.1 (Goursat’ Lemma/Goursat’s Theorem) If U is an open set of C andf : U → C is holomorphic, then for any triangle T , whose interior and the triangleitself is contained in U , ∫

T

f(z)dz = 0.

37

Figure 4.2: Graph by E. Hairer and G. Wanner

Proof We may assume that U is connected. Denote T (0) = T . Let us take the middlepoint on each sides of T (0) and obtain four triangles covering T whose vertices are thevertices of T (0) and the middle points, which we denote by T (1)

i , i = 1, 2, 3, 4 and∫T (0)

f(z)dz =

4∑i=1

∫T

(1)i

f(z)dz.

The extra sides of the smaller triangles are traversed twice, in opposite directions andhence the integral of f along them canceled out. Let T (1) denote one of the fourtriangles s.t. |

∫T (1) f(z)dz| is the largest among the four values |

∫T

(1)if(z)dz|. We

have ∣∣∣∣∫T

f(z)dz

∣∣∣∣ ≤ 4

∣∣∣∣∫T (1)

f(z)dz

∣∣∣∣ .We continue with this procedure, divide T (1) to four triangles from which we select

a triangle T (2). In this way we obtain a sequence of triangles T (i) whose enclosedregions are nested and∣∣∣∣∫

T

f(z)dz

∣∣∣∣ ≤ 4

∣∣∣∣∫T (1)

f(z)dz

∣∣∣∣ ≤ · · · ≤ 4n∣∣∣∣∫T (n)

f(z)dz

∣∣∣∣ .Note that

T ⊃ T (1) ⊃ T (2) ⊃ . . . .

The sequence of nested triangles contains a common point which we denote by z0. Thelength between z0 and any other points in the triangle T (n) is less than 2−nL where Lis perimeter of the original triangle. Also the side of T (n) is also less than 2−nL.

Since f is holomorphic there exists a function ψ with limz→z0 ψ(z) = 0 and

f(z) = f(z0) + f ′(z0)(z − z0) + ψ(z)|z − z0|.

38

Since f(z0)+f ′(z0)(z−z0) has a primitive its integral along any closed curve vanishes.Hence ∫

T (n)

f(z)dz =

∫T (n)

ψ(z)|z − z0|dz.

Since |z − z0| ≤ 2−nL and

4n∣∣∣∣∫T (n)

f(z)dz

∣∣∣∣ ≤ 4nlength(T (n))2−nL maxz∈T (n)

|ψ(z)| ≤ 4n(2−nL)2 maxz∈T (n)

|ψ(z)| → 0.

Consequently,∫Tf(z)dz = 0. �

Goursat (1884) actually proved the above for rectangles by sub-dividing rectangles.

Corollary 4.4.2 Suppose that f is holomorphic in a region U , then for any rectangleR who and whose interior contained entirely in U ,∫

R

f(z)dz = 0

Proof Just divide R into two triangles T1 and T2 whose share a side which are trans-versed twice in opposite directions. Hence∫

R

f(z)dz =

∫T1

f(z)dz +

∫T2

f(z)dz.

�

A similar proof shows that in Goursat’s theorem we may replace the triangle byany polygons. Argue by induction on the number of sides in a polygon, once see thatGoursat’s theorem holds for polygons.

Theorem 4.4.3 (Cauchy’s Theorem, existence of primitive) A holomorphic functionin a star region U has

F (z) =

∫[c,z]

f(ζ)dζ, z ∈ U

as its primitive, where C is a centre of U . If γ is a closed curve in U then∫γ

f(z)dz = 0.

Proof By Goursat’s Lemma,∫Tf(z)dz = 0 for any triangles in U . By Theorem 4.3.3

on the existence of primitives in a star region, f has a primitive F given by the statedformula. Apply Theorem 4.3.1 to conclude the vanishing of the curve integral. �

4.5 Cauchy’s Theorem: simply connected domains (Lec-ture 12)

Definition 4.5.1 A connected set is simply connected if any continuous curve can becontinuously deformed into a point.

39

The precise meaning of the continuous deformation is the following: there exists acontinuous map F (s, t) : [0, 1] × [a, b] → C such that F (0, ·) : [a, b] → C is aparameterization for γ, F (1, ·) : [a, b] → C is a constant curve, i.e. F (1, t) = F (1, 0)for all t.

Remark 4.5.1 A star region is simply connected. Let γ1, γ2 be two continuous curveswith the same end points. Let z : [a, b] → C be a parameterization of γ1. Then a lineconnecting C to z(t) pass through a point on γ2 which we denote by z(t). Set

F (s) = (1− s)z(t) + sz(t).

Remark 4.5.2 A region in C is simply connected if and only if its complement isconnected in the extended plane C∗. A bounded region in C is simply connected if andonly if its complement is connected in C.

The strip {x+ iy : 1 < x < 2} is simply connected. Whenever a closed simple curvelies in a simply connected region, its interior lies also in the region. If U is a region, itis easy to spot a subset of C \ U that is a positive distance from the rest. Such subsetsare ‘holes’ in U .

Theorem 4.5.3 (Cauchy’s Theorem, c.f. Theorem 4.4.3) If f is a holomorphic func-tion in a simply connected domain U then∫

γ

f(z)dz = 0

for any closed curve in U .

The proof for this theorem is in Section 4.6. For now we do not prove this theorem, andhence will limit the use of Cauchy’s Theorem for simply connected domain to enhanceunderstanding.

The following are non-simply connected regions:

{z : r1 < |z − z0| < r2}, {z : 0 < |z − z0| < r2}

Also, not simply connected:

{|z| < 20} \ {{|z − i| < 2} ∪ {|z − 1| < 1}}.

We are forced to consider non simply connected domains if a function has singularities.

Definition 4.5.2 A point z0 is an isolated singularity of f if f is holomorphic in somepunctured disk {0 < |z − z0| < r}, not differentiable or not defined at z0.

The function f(z) = 1z has an isolated singularity at 0.

Example 4.5.4 The function 1sin z has singularities at nπ where n ∈ Z.

For this we need to check the function sin(z) = 0 at and only at z = nπ. Firstlyusing the formula sin(z) = eiz−e−iz

2i we see that

sin(x+ iy) = sinx cosh y + i cosx sinh y

40

where cosh z = ez+e−z

2 and sinh z = ez−e−z2 . So sin(z) vanishes if and only if

sinx cosh y = 0, cosx sinh y = 0.

Since cosh y > 0 for y ∈ R, we have sinx = 0 and x = nπ. For x = nπ,cosx sinh y = ± sinh y which vanishes only if y = 0. Thus the zero’s of sin z arenπ.

Example 4.5.5 f(z) =√z does not have an isolated singularity at 0. It cannot be

defined on any punctured disc around 0.

In the next chapter we discuss the integral of a function f along a contour curvewhich enclose an isolated singularity of f . We learn how to divide a circle or more gen-erally a star region with punctured holes into star regions / simply connected regions.

4.6 SupplementaryWe give a proof for Cauchy’s theorem for simply connected regions.

Theorem 4.6.1 Let U be a simply connected region. Let f : U → C be holomorphic.If γ1 and γ2 are homotopic then∫

γ1

f(z)dz =

∫γ2

f(z)dz.

Proof Let F : [0, 1] × [a, b] → U be a continuous map such that F (0, ·) is a parame-terization for γ1 and F (1, ·) a parameterization for γ2.

Suppose that s1 − s2 is sufficiently small, the two curves γ1 = F (s1, ·) and γ2 =F2(s2, ·) are close to each other:

maxa≤t≤b

|F (s1, t)− F (s2, t)| ≤ ε.

Then there exists a finite division of [a, b], a = t0 < t1 < · · · < tn+1 = b suchthat γ1 is the union of a finite number of curves γ(k)

1 (t), t ∈ [tk, tk+1]. Similarly γ2 isthe union of γ(k)

2 (t), t ∈ [tk, tk+1]. Then it is clear∫γi

F (z)dz =

n∑k=0

∫γ(k)i

F (z)dz, i = 1, 2.

We connect γ(k)1 with γ(k)

2 with two line segments to obtain a closed curve Γk. Wechoose ε small so each region enclosed by Γk lie entirely in a disc contained in U .Then by Cauchy’s theorem in a disc,∫

Γk

f(z)dz = 0.

Summing up over k, we seen∑k=0

∫γ(k)1

F (z)dz =

n∑k=0

∫γ(k)2

F (z)dz.

This completes the proof.�

41

Chapter 5

Cauchy’s Integral Formula

If f is holomorphic in a region U then f has derivatives of all orders. Furthermore if z0

is such that a disc D centred at z0 is contained entirely in U , then it has a power seriesexpansion around z0 on D. Furthermore,

f (n)(z0) =n!

2πi

∫|z−z0|=r

f(ζ)

(ζ − z0)n+1dζ.

This is intimately related to the properties of the function 1z−z0 .

5.1 Keyhole Operation and Other Techniques (Lecture12)

In this section we learn how to operate with domains with holes and in doing so weprove a simpler form of the Cauchy integral formula whose general form will be provedusing a different technique later.

It is much easier to describe a curve as the boundary of a region (the contour). Theboundary curve of a region may not be very nice. We always assume that the curvesis piecewise smooth and oriented positively, so the the enclosed region stays on its leftas we travel in the positive direction along the boundary. If U is a region the boundarycurve of U is denoted by ∂U .

One of the techniques is called the key hole operation.

D(z0, r) = {z : |z − z0| < r}, C(z0, r) = {z : |z − z0| = r}.

Lemma 5.1.1 (Keyhole Lemma) Let γ be the piecewise smooth and positively ori-ented boundary of a star region U . Suppose that z0 ∈ U and f is holomorphic at everypoint of an open set containing U with the exception z0. Let r be sufficiently small sothe closed disc centred at z0 of radius r is contained in U . Then∫

γ

f(z)dz =

∫C(z0,r)

f(z)dz.

We give two proofs, representing two techniques.Proof This technique is called key hole operation. Denote C the circle |z − z0| = r.Let us consider the line from the centre of U to z0 whose continuation intersects with

42

γ at a point on the boundary which we denote by a. Take any line through z0 if thecentre coincide with z0. Then U = U \ {[z0, a]} is a star region with the same centre,on which f is holomorphic.

We take a narrow corridor of width ε along the segment [z0, a], connecting to thecircle C. This gives a closed path Γε contained in U . We now apply Cauchy’s theoremto f on U and get

∫Γεf(z)dz = 0. Since f(z) is bounded, the sum of the integrals

of f(z) along the two sides of the corridors vanishes, as ε → 0. Its integral along themissing arc of γ vanishes as ε converges to 0. Thus

∫γf(z)dz =

∫Cf(z)dz as stated.

�

Proof This proof is based on Cauchy’s theorem for simply connected domains. Theline segment passing through the centre of U and z0, (take any line through z0 if thecentre coincide with z0), splits the region U \ {|z− z0| = r} into two parts. Let γ1 andγ2 denote the boundaries of the two parts (each curve contains one part of C and onepart of γ). Then γ1 (also γ2) is contained in a simply connected domain in which f(z)is holomorphic. By Cauchy’s theorem for simply connected domains,∫

γi

f(z)dz = 0, i = 1, 2.

Also ∫γ

f(z)dz −∫C

f(z)dz =

∫γ1

f(z)dz +

∫γ2

f(z)dz = 0.

�

Corollary 5.1.2 Let U be a star region containing z0 with piecewise smooth boundaryγ, f a holomorphic function in an open set containing U , then∫

γ

1

z − z0dz = 2πi.

Also for n 6= −1, ∫γ

1

(z − z0)ndz = 0.

Proof For any r,∫|z−z0|=r

1

z − z0dz =

∫ 2π

0

1

reitd

dt(z0 + reit)dt = i

∫ 2π

0

dt = 2πi.

Also for an integer n 6= 1,∫|z−z0|=r

1

(z − z0)ndz =

∫ 2π

0

1

rneitnd

dt(z0 + reit)dt = ir1−n

∫ 2π

0

1

eit(n−1)dt

= ir1−n∫ 2π

0

(cos((1− n)t) + i sin((1− n)t)) dt = 0.

�

43

5.2 Cauchy’s Integral Formula

Theorem 5.2.1 (Cauchy’s Integral formula) Suppose U is a star region with piece-wise smooth positively oriented smooth boundary γ. Suppose f is holomorphic in anopen set containing U . Then for every z0 in U ,

f(z0) =1

2πi

∫γ

f(z)

z − z0dz.

Proof Let Cε = {|z − z0| = ε} where ε is sufficiently small. Then by Lemma 5.1.1,∫γ

f(z)

z − z0dz =

∫Cε

f(z)

z − z0dz.

Now ∫Cε

f(z)

z − z0dz =

∫Cε

f(z)− f(z0)

z − z0dz + f(z0)

∫Cε

1

z − z0dz

=

∫Cε

f(z)− f(z0)

z − z0dz + f(z0)2πi.

Since f is holomorphic, f(z)−f(z0)z−z0 is bounded by a number M on a disc,∣∣∣∣∫

Cε

f(z)− f(z0)

z − z0dz

∣∣∣∣ ≤M2πε→ 0, as ε→ 0.

Finally ∫γ

f(z)

z − z0dz = lim

ε→0

∫Cε

f(z)− f(z0)

z − z0dz + f(z0)2πi = f(z0)2πi,

completing the proof. �

Corollary 5.2.2 (Cauchy’s Integral formula for a disc) If f is holomorphic in an openset containing the disc D = {z : |z − z0| ≤ r} then for every z in {z : |z − z0| < r},

f(z) =1

2πi

∫C

f(ζ)

ζ − zdζ,

where C = {z : |z − z0| = r} is the boundary circle, positively oriented.

The corollary says that the value of a holomorphic function in a disc is determined byits values on the boundary circle. In particular,

f(z0) =1

2πi

∫C

f(ζ)

ζ − z0dζ =

1

2π

∫ 2π

0

f(z0 + reit)dt.

44

5.3 Taylor Expansion, Cauchy’s Derivative FormulasWe prove a holomorphic function has a power series expansion around every point, andis hence analytic. There are integral formulas for its derivatives.

Theorem 5.3.1 (Cauchy’s Theorem on Taylor expansion) If f is holomorphic on anopen disc D(z0, R) = {z : |z − z0| < R}, then there exists a sequence of complexnumbers an such that

f(z) =

∞∑n=0

an(z − z0)n, z ∈ D(z0, R).

The an’s are given by

an =1

2πi

∫C

f(z)

(z − z0)n+1dz,

where C is a circle centred at z0 with radius r < R. In particular f is infinite timesdifferentiable and is analytic on D(z0, R).

Proof Let z ∈ D(z0, R). Choose a number r between |z − z0| and R. By Cauchy’sintegral formula, Corollary 5.2.2,

f(z) =1

2πi

∫|ζ−z0|=r

f(ζ)

ζ − zdζ.

Now1

ζ − z=

1

(ζ − z0)− (z − z0)=

1

ζ − z0

1

1− z−z0ζ−z0

.

Since∣∣∣ z−z0ζ−z0

∣∣∣ = |z−z0|r < 1,

f(ζ)

ζ − z=

f(ζ)

(ζ − z0)

∞∑n=0

(z − z0)n

(ζ − z0)n.

On |ζ − z0| = r,∣∣∣∣ f(ζ)

(ζ − z0)

(z − z0)n

(ζ − z0)n

∣∣∣∣ ≤ max|ζ−z0|≤r |f(ζ)|r

|z − z0|n

rn.

Since∑∞n=0

|z−z0|nrn < ∞, by the dominated convergence theorem, we may integrate

term by term,

f(z) =1

2πi

∫|ζ−z0|=r

f(ζ)

ζ − zdζ =

∞∑n=0

(z − z0)n1

2πi

∫|ζ−z0|=r

f(ζ)

(ζ − z0)n+1dζ.

Then f(z) =∑∞n=0 an(z − z0)n. By Lemma 5.1.1, the value∫

|ζ−z0|=r

f(ζ)

(ζ − z0)n+1dζ

makes sense for all 0 < r < R and is independent of r. By Theorem 3.2.2, a powerseries function is analytic in its disc of convergence . �

45

Remark 5.3.2 Note if f(z) =∑∞n=0 an(z − z0)n then an = f(n)(z0)

n! .

By the theorem above if f is holomorphic on a disc around z0 it has a power seriesexpansion whose disc of convergence, if not the whole complex plane, must meet apoint where f is not differentiable (singularity). If f is holomorphic on D(z0, R), itspower series expansion has radius of convergence R.

Corollary 5.3.3 Let f be a holomorphic function on D(z0, R). Let z ∈ D and r besuch that |z − z0|+ r < R, then

f (n)(z) =n!

2πi

∫|ζ−z|=r

f(ζ)

(ζ − z)n+1dζ.

Proof By the standard theory on power series, and the last paragraph in the proof wesee that f is infinitely times differentiable. And, (using for example Theorem 3.1.2),

f (n)(z) =n!

2πi

∫|ζ−z|=r

f(ζ)

(ζ − z)n+1dζ,

where r is a number such that {|ζ − z| ≤ r} is contained in the disc D(z0, R), con-cluding the formula. �

Under the assumption of the Corollary, if |z − z0| < δ < R, the following alsoholds:

f (n)(z) =n!

2πi

∫|ζ−z0|=δ

f(ζ)

(ζ − z)n+1dζ.

Indeed, as |z − z0| < δ, D(z0, r) contains z, we may use Lemma 5.1.1 to change thecontour curve, to obtain∫

|ζ−z|=r

f(ζ)

(ζ − z)n+1dζ =

∫|ζ−z0|=δ

f(ζ)

(ζ − z)n+1dζ.

Theorem 5.3.4 If f is holomorphic in an open set U , then f is infinitely times differ-entiable and is analytic. If C ⊂ U is a circle whose interior is also contained in Uthen

f (n)(z) =n!

2πi

∫C

f(ζ)

(ζ − z)n+1dζ

for all z in the interior of C.

Proof For each z0 ∈ U , simply apply the earlier theorem to a small disc D(z0, R) thatis contained in U . �

5.4 Estimates, Liouville’s Thm and Morera’s ThmDenote by C(z0, R) the circle centred at z0 of radius R.

Theorem 5.4.1 (Cauchy’s Inequality) If f is holomorphic in an open set U and if theclosed disc {z : |z − z0| ≤ R} is contained in U , then

|f (n)(z0)| ≤ supz∈C(Z0,R)

|f(z)| · n!

Rn.

46

Also if f(z) =∑∞n=0 an(z− z0)n on an open set containing the closed disc D(z0, R),

then|an| ≤

1

Rn· supz∈C(Z0,R)

|f(z)|.

Proof By Cauchy’s derivative formula,

f (n)(z0) =n!

2πi

∫C(z0,R)

f(ζ)

(ζ − z0)n+1dζ.

Taking the parameterization, Reit,

|f (n)(z0) ≤ n!

2π

∣∣∣∣∫ 2π

0

f(z0 +Reit)

(Reit)n+1iReitdt

∣∣∣∣≤ n!

2πsup

t∈[0,2π]

|f(z0 +Reit)|∫ 2π

0

1

Rndt = sup

z∈C(Z0,R)

|f(z)| · n!

Rn.

The second part can be proved similarly,

|an| =∣∣∣∣ 1

2πi

∫C

f(z)

(z − z0)n+1dz

∣∣∣∣ ≤ supz∈C(Z0,R)

|f(z)| · 1

Rn.

�

Corollary 5.4.2 (Liouville’s Theorem) A bounded entire function is a constant.

Proof Let |f |∞ = supz∈C |f(z)|. By Cauchy’s inequality, for any z0, and any R,

|f ′(z0)| ≤ |f |∞1

R.

Taking R to infinity to see that f ′(z0) = 0. Since z0 is an arbitrary point in C, f ′

vanishes identically and f is a constant (by Corollary 4.3.2). �

Example 5.4.3 Since sin(z) is an entire function, by Liouville’s theorem it cannot bebounded on C, a fact can also easily be deduced from the formula sin(x + iy) =

sinx ey+e−y

2 + i cosx ey−e−y

2 .

Corollary 5.4.4 (Morera’s Theorem) If f is continuous in an open disc D s.t.∫T

f(ζ)dζ = 0

for any triangle T contained in this disc, then f is holomorphic on D.

Proof By the integrability criterion, Theorem 4.3.3, f has a primitive F in D. ByCauchy’s theorem for Taylor expansions, F is infinite many times differentiable, andso in particular f = F ′ is holomorphic. �

47

5.4.1 SupplementaryAn extension to Liouville’s theorem is:

Example 5.4.5 If f is an entire function with limz→∞f(z)z = 0, then f is a constant.

Proof Let z0 ∈ C, R > 0. By Cauchy’s derivative formula,

f ′(z0) =1

2πi

∫|ζ−z0|=R

f(ζ)

(ζ − z0)2dζ =

1

2π

∫ 2π

0

f(z0 +Reit)

Reitdt.

Hence

|f ′(z0)| ≤ 1

2π

∣∣∣∣∫ 2π

0

f(z0 +Reit)

z0 +Reitz0 +Reit

Reitdt

∣∣∣∣ ≤ sup0≤t≤2π

∣∣∣∣f(z0 +Reit)

z0 +Reit

∣∣∣∣ |z0|+R

R.

The right hand side converges to 0 as R → ∞. Hence f ′ = 0 identically and f is aconstant. �

Example 5.4.6 Let f be an entire function such that there exist real numbersR andMsuch that |f(z)| ≤M |z|n outside of the disc {z : |z| ≤ R}. Then f is a polynomial ofdegree at most n.

Proof An entire function f has a power series expansion at 0:

f(z) =

∞∑n=0

anzn, z ∈ C.

For m > n, and for any r > R,

|am| =1

2πi

∣∣∣∣∣∫|z|=r

f(z)

zm+1dz

∣∣∣∣∣ =1

2πi

∣∣∣∣∣∫|z|=r

f(z)

zn1

zm+1−n dz

∣∣∣∣∣≤ sup|z|=r

|f(z)||zn|

1

rm+1−n (2πr)→ 0,

as r →∞. Thus am = 0 for all m ≥ n and f is a polynomial of at most degree n. �

5.5 Locally Uniform Convergent Sequence of Holomor-phic Functions

The limit of a sequence of continuous functions, converging uniformly, is continuous.Continuity is a concept which is the same weather we treat f as a complex valuedfunction of one complex variable or as a R2 valued function of two real variables.

Theorem 5.5.1 (The Uniform Convergence Theorem/Weierstrass Theorem) Let fnbe a sequence of holomorphic function on an open set U . Suppose that fn convergesuniformly to a function f on any compact subsets of U , then f is holomorphic. Fur-thermore f ′n converges to f ′ uniformly on compact subsets of U .

48

Proof Part (1). Let z0 ∈ U and D = D(z0, R) a disc with its closure containedentirely in U . By Goursat’s lemma for holomorphic functions on a disc, for each n,∫

T

fn(z)dz = 0

for any triangle in D. Since fn → f uniformly on D, f is continuous on D and∫T

fn(z)dz →∫T

f(z)dz

and the latter vanishes. By Morera’s theorem, f is holomorphic on D(z0, R). Since z0

is an arbitrary point in U , we may conclude that f is holomorphic.Part (2). Let K be a compact subset of U . Then there exists R > 0 such that for

each z0 ∈ K, the disc D(z0, R) is contained in U , and such that for any r < R, K canbe covered by a finite number of ball of radius r. To prove f ′n → f ′ uniformly on K,we only need to prove f ′n converges uniformly for every discD(z0, r) where r < R/2.

Let z0 ∈ U , we prove that f ′n → f ′ uniformly on D(z0, r) where 2r < R andD(z0, R) ⊂ U . Let

C = {z : |z − z0| = 2r}.

By Cauchy’s derivative formula, for any z in the interior of C,

f ′n(z)− f ′(z) =1

2πi

∫C

fn(ζ)− f(ζ)

(ζ − z0)2dζ.

For any ε > 0 there exists N s.t. for n > N ,

supz∈D(z0,2r)

|fn(z)− f(z)| ≤ εr.

Hence for n > N and z ∈ {|z − z0| ≤ r},

|f ′n(z)− f ′(z)| ≤ 1

2π

εr

(2r)2(4πr) ≤ ε,

where (4πr) is the length of the circle C ( c.f. Theorem 4.2.2). We have proved thatf ′n → f ′ uniformly on compact sets.

�

5.6 Schwartz Reflection Principle (Lecture 15)This section can be considered as an application of Morera’s Theorem. Schwartz Re-flection Principle is a method of extending a holomorphic function over and to the otherside of the real axis.

Let U be an open subset of C, symmetric with respect to the real axis, i.e.

z ∈ U iff z ∈ U.

Let I = U ∩ R,

U+ = {z ∈ U, Im(z) > 0}, U− = {z ∈ U, Im(z) < 0}.

Then U = U+ ∪ I ∪ U−.

49

Theorem 5.6.1 (Symmetry Principle) If f+ and f− are respectively holomorphicfunctions on U+ and U−, that extends continuously to I , and f+(x) = f−(x) forany x ∈ I , define

f(z) =

f+(z), z ∈ U+

f+(z) = f−(z), z ∈ If−(z), z ∈ U−.

Then f is holomorphic on U = U+ ∪ I ∪ U−.

Proof It is clear that f is continuous on U , holomorphic on U+ and on U−. Let z0 ∈ Iand D a disc in U centred at z0. Let T be a triangle in D we prove that

∫Tf(z)dz = 0.

There are three possibilities:

(a) T lies entirely within U+ or entirely within U−, then∫Tf(z)dz = 0 by Gour-

sat’s lemma.

(b) One side of T is on I . For example assume T is in U+ ∪ I , in this case we take atriangle Tε by moving its side on I upward, horizontally, with its distance fromI to be ε. Then Tε → T and∫

Tε

f(z)dz →∫T

f(z)dz = 0.

By part (a),∫Tεf(z)dz vanishes for any ε, hence

∫Tf(z)dz = 0.

(c) T lies cross I in this case we may divide the triangles, T1 lies entirely on side ofthe I axis, each of the other triangles T2 and T3 have one side on I . By part (a)and (b),

∫Tif(z)dz = 0 for i = 1, 2, 3 and

∫T

f(z)dz =

3∑i=1

∫T

f(z)dz = 0.

We apply Morera’s theorem to conclude that f is holomorphic on D. Since z0 is anaribitary point in I , f is holomorphic at each point of I . �

The following Lemma is given in Example Sheet 1.

Theorem 5.6.2 (Schwartz Reflection Principle) Suppose f is a holomorphic func-tion on U+ extending continuously to I , and f takes real value on I . Then there existsa holomorphic function F on U such that F and f agree on U+ ∪ I .

Proof We define

F (z) =

f(z), z ∈ U+ ∪ I

f(z) z ∈ U−.

For z ∈ I , F (z) = F (z) and so the function is well defined. We have seen that ifg : U− → C given by g(z) = f(z) is holomorphic. By the reflection principle, F isholomorphic on U .

�

There are two proof for the complex differentiability of g : U− → C given byg(z) = f(z) where f is holomorphic on U+. Proof 1. (On Example sheet 1) Let

50

z0 ∈ U− and z ∈ U−. Since f is differentiable in z0, there exists a function ψ withlimζ→z0 ψ(ζ) = 0 and

f(z) = f(z0) + f ′(z0)(z − z0) + ψ(z)|z − z0|.

Taking conjugate of both side we see that

g(z) = g(z0) + f ′(z0)(z − z0) + ψ(z)|z − z0|.

Since limz→z0 ψ(z) = 0 this proves that g is complex differentiable at z0 and g′(z0) =f ′(z0).

Proof 2. Let z0 ∈ U−. We make a Taylor expansion of f at z0 around a disc Dcontained in U+:

f(z) =

∞∑n=0

an(z − z0)n.

Then taking conjugate of both sides we see

g(z) =

∞∑n=0

an(z − z0)n,

hence g is analytic on U−.

5.7 The Fundamental Theorem of AlgebraThe zero of a function f is a solution to f(z) = 0.

Lemma 5.7.1 Every non-constant polynomial p(z) = a0 + a1z + · · · + anzn has at

least one zero.

Proof Suppose that p(z) has no root then 1p(z) is an entire function. We only need to

prove it is bounded. A bonded entire function is a constant.The entire function 1

p(z) is certainly bounded on any compact set. At far away itconverges to zero and so is also bounded on C \K where K is a compact set.

Detail for boundedness of 1p(z) . Suppose an 6= 0 for n ≥ 1. If z 6= 0,

p(z)

zn= (

a0

zn+

a1

zn−1+ · · ·+ an−1

z) + an.

The first term converges to 0 as z →∞ and∣∣∣p(z)zn

∣∣∣ ≥ |an|2 for |z| sufficiently large and∣∣∣∣ 1

p(z)

∣∣∣∣ =|zn||p(z)|

1

|z|n≤ 2

|an|1

|z|n,

which converges to zero as z →∞. Hence 1p(z) is a bounded. �

Theorem 5.7.2 (The Fundamental Theorem of Algebra) Every polynomial p(z) =a0 + a1z + · · · + anz

n of degree n ≥ 1 has precisely n zeros in C. Furthermore ifz1, . . . , zn are the roots then

p(z) = an(z − z1)(z − z2) . . . (z − zn).

51

Proof By the previous lemma p(z) = 0 has at least one root, z1. Write z = (z−z1)+z1

and expand p(z) in z − z1. There exist b0, b1, . . . , bn such that

p(z) = b0 + b1(z − z1) + · · ·+ bn(z − z1)n.

Evaluate p(z) at z1 giving b0 = 0. Hence

p(z) = (z − z1)(b1 + b2(z − z1) + . . . bn(z − z1)n−1 = (z − z1)Q(z).

By induction on the degree of the polynomial, Q(z) = 0 has n− 1 roots and

p(z) = A(z − z1) . . . (z − zn).

The number A is the coefficients in front of zn and is hence an. �

5.8 Zero’s of Analytic FunctionsZero’s of analytic functions are useful for a number of reasons: they are closely relatedto poles of meromorphic functions; the uniqueness of analytic continuations, and thedistributions of eigenvalues of a dynamical system. Or simply one wants to count thezero’s, e.g. that of the zeta function!

The humblest application is to know where are the zero’s of the derivative of aholomorphic function, which is also a holomorphic function. c.f. The Inverse FunctionTheorem and Conformal Mappings.

Let pn(z) be a polynomial of degree n and z0 a zero of pn(z). Then there exists anatural number r and w1, . . . , wn−r, not equal to z0 such that

pn(z) = an(z − z0)r(z − w1) . . . (z − wn−r).

The holomorphic function h(z) = an(z−w1) . . . (z−wn−r) does not vanish in a disccontaining z0. Furthermore,

p′n(z) = pn(z)

(r

z − z0+

1

z − w1+ · · ·+ 1

z − wn−r

).

If r > 1 it is clear p′n(z0) = 0. If r = 1,

p′n(z) =

(anΠn−1

i=1 (z − wi) + pn(z)

(1

z − w1+ · · ·+ 1

z − wn−1

)),

and p′n(z0) = anΠn−1i=1 (z0 − wi) 6= 0.

By induction on r we conclude that if z0 is a zero, or order r, of pn(z) then thefollowing holds:

• For k = 0, 1, . . . , r − 1, p(k)n (z0) = 0 (the k th derivatives).

• p(r)n (z0) 6= 0.

• There exists a holomorphic function h such that h(z) 6= 0 on a discD containingz0 and

pn(z) = (z − z0)rh(z), z ∈ D.

52

Definition 5.8.1 Let f be a holomorphic function on a region U . The order of a zeroz0 of f is defined to be:

ord(f, z0) = inf{k : f (k)(z0) 6= 0}.

A trivial function is one which vanishes identically. A non-vanishing function is afunction that does not vanish anywhere.

We prove a multiplicative statement for a general holomorphic function.

Lemma 5.8.1 (Multiplicative Statement) Suppose that f is a holomorphic functionin a region U with a zero at z0. There exists a neighbourhood D of z0 such that eitherf is identically zero on D or there exists a unique non-vanishing holomorphic functionh on D, and a unique natural number k such that

f(z) = (z − z0)kh(z), ∀z ∈ D.

This number k is the order of z0.

Proof (1) Since f is holomorphic, it has a power series expansion at z0, on a discD = D(z0, r):

f(z) =

∞∑n=0

an(z − z0)n, ∀z ∈ D.

Firstly a0 = f(z0) = 0 by the assumption. Suppose that f does not vanish identicallyon D. Let k ≥ 1 be the first non-zero term in the power series expansion. Then

f(z) = (z − z0)kak

(1 + (z − z0)

∞∑n=k+1

anak

(z − z0)n−k−1

).

Define

h(z) = ak

(1 + (z − z0)

∞∑n=k+1

anak

(z − z0)n−k−1

).

There exists M such that

sup|z|≤r/2

∣∣∣∣∣∞∑

n=k+1

an(z − z0)n−k−1

∣∣∣∣∣ ≤M,

as the power series inside the modulus sign converges on D(z0, r/2) and is a con-tinuous function on the disc D. Let r = min(r/2, 1

2M ). Then h is holomorphicand does not vanish on the open disc {|z − z0| ≤ r}. (3) Uniqueness. Supposef(z) = (z − z0)kh(z) = (z − z0)k h(z). If k > k then h(z) = (z − z0)k−k h(z) andh(z0) = 0, which contradicts with the definition of h. Consequently, k = k, and alsoh = h.

(4) Finally, f ′(z) = k(z − z0)k−1h(z) + (z − z0)kh′(z). If k = 1, f ′(z0) =h(z0) 6= 0. If k > 1, f ′(z0) = 0. Induction on the degree k,

f (j)(z) = k(k − 1) . . . (k − j + 1)(z − z0)k−jh(z) + (z − z0)k−j+1G(z)

for a holomorphic function G. Thus f (j)(z0) = 0 for any j < k and f (k)(z0) 6= 0. �

53

Theorem 5.8.2 If f is a holomorphic function with f(z0) = 0. Then there exists adisc D(z0, r) on which either f vanishes identically or f has no other zero’s.

Proof Suppose that f(z) = (z − z0)kh(z) on {|z − z0| < r} where h is holomorphicand does not vanish. Then f(z) does not vanish on {z : 0 < |z − z0| < r}. �

Definition 5.8.2 A zero of f , z0, is said to be isolated if there exists an open disc Daround z0 such that f(z) 6= 0 for any z ∈ D \ z0.

Every zero of a holomorphic function is isolated unless f is trivial on a disc near z0.

Theorem 5.8.3 Let U be a region in C.

• Suppose g1 and g2 are holomorphic in U and g1, g2 agree on a sequence of zk inU who has an accumulation point z0 in U and zk 6= z0 for any k. Then g1 = g2

identically on U .

• In particular, if f is a holomorphic function in U with a sequence of zeros’szk ∈ U converging to z0 in U , and z0 6= zk for any k. Then f(z) is identically 0on U .

Proof Setting f = g1 − g2 we only need to prove the special case. By the continuity,f(z0) = limk→∞ f(zk) = 0. Within every disc D(z0,

1n ) f has a zero other than z0.

By Theorem 5.8.2, f ≡ 0 on a disc around z0. We proceed to prove that f is identicallyzero on U .

Let V be the interior of the set of zeros of f . Then V is non empty, it contains z0,and is an open set by the construction. Let zn ∈ V → z0. Then f(z0) = 0. Sincez0 is the accumulation point of distinct zero’s, f vanishes in a disc around z0 and z0 iscontained in V . Hence z0 ∈ V . This proves V is closed.

Thus U is the union of two open sets:

U = (U \ V ) ∪ V.

If U \V is not empty, U is the disjoint union of nonempty open sets, contradicting withthe fact that U is connected. If U \ V is empty, then f is identically zero.

�

The multiplicative statement can now be strengthened as below.

Theorem 5.8.4 (Multiplicative Statement) Suppose that f is a non-trivial holomor-phic function in a region U and has a zero z0. There exists a neighbourhood D of z0,a unique non-vanishing holomorphic function h on D, and a unique natural number ksuch that

f(z) = (z − z0)kh(z), ∀z ∈ D.

This number k is the order of z0.

Proof By the Lemma 5.8.1, there exists a disc D around z0 on which either f van-ishes or f(z) = (z − z0)kh(z). The former implies that f vanishes identically on Ucontradicting the assumption f is not trivial (i.e. not everywhere 0). �

54

5.9 Uniqueness of Analytic Continuation (Lecture 17)Let γ be a simple closed curve and U the interior region determined (enclosed in ) byγ. Let f : U → C be an analytic function in U . It is an interesting question whether fca be analytically continued from a boundary point.

Definition 5.9.1 Suppose f is holomorphic on U . Let z0 be a point on the boundaryof U . We say f is analytically continuable at z0 if there exists a function g : U ′ → Rwhere U ′ is an open set containing z0 such that g and f agree in U ∩ U ′.

Corollary 5.9.1 Suppose that f1, f2 are defined and holomorphic on U1 and U2 re-spectively. If U1 ∩ U2 is connected and f1 = f2 on a sequence of point zk ∈ U1 ∩ U2

with a limit z0 s.t. zk 6= z0 for any k then f1 = f2 on U1 ∩ U2.

Corollary 5.9.2 Let f be a holomorphic function in a region U . Let V ⊂ U be aclosed bounded subset of U . Then for any w ∈ C, {z ∈ V : f(z) = w} has a finitenumber of elements or f is a constant on V .

Proof Let g(z) = f(z) − w. For any z ∈ V there exists a number r(z) > 0 suchthat w is not in the image of D′(z, r(z)) or f(z) ≡ w on this disc, in which case f isidentically w on U . Since V is compact there exists a finite number of points z, . . . , zmin V such that

V ⊂ ∪mi=1D′(zi, zi(r)).

It is clear that {z ∈ V : f(z) = w} ⊂ {z1, . . . , zm}. �

5.10 SupplementaryThe Lacunary function is:

f(z) =

∞∑n=1

z2n = z + z2 + z4 + z8 + . . . .

The radius of convergence of the power series is R = 1. The largest open set on whichit can be extended analytically is the unit disc. In fact it has singularity at all 2n throots of the unity. Since the singularity points on the unit circle is dense in the circle,it cannot be analytically continued beyond the disc.

55

Chapter 6

Laurent Series and Singularities

A series of the formb1z−1 + b2z

−2 + . . .

can be considered as a power series in 1z . Let R denote its radius of convergence R. If

R is finite number, the power series converge absolutely for z with |z| > 1R . As usual,

1∞ is interpreted as 0.

If this series is combined with an ordinary power series we have a general series ofthe form

∞∑n=−∞

anzn.

Definition 6.0.1 The series is said to converge, if both its positive power part∑∞n=0 anz

n

and negative power part∑−∞n=−1 anz

n are separately convergent.

Suppose that the positive part of the power series has a radius of convergence R2 > 0and the negative part of the power series converges when |z| > R1. If R1 < R2,they have a common region of convergence: R1 < |z| < R2. Conversely if f isholomorphic function in a region contains the annulus R1 < |z − z0| < R2, it has ageneral series expansion at z0:

f(z) =

∞∑n=−∞

an(z − z0)n.

This is called a Laurent series (development) for f .

6.1 Laurent Series Development (Lecture 17-18)Lemma 6.1.1 Let f be a holomorphic function on a disc D(z0, r). Define

g(z) =

{f(z)−f(z0)

z−z0 , z 6= z0

f ′(z0), z = z0.

Then g is holomorphic on D(z0, r).

56

Proof As the quotient of two holomorphic functions, g is differentiable on D \ {z0}.We only need to prove g is differentiable at z0. Firstly, there is a Taylor series expansionfor the holomorphic function f in a neighbourhood of z0

f(z) = f(z0) + f ′(z0)(z − z0) +

∞∑n=2

an(z − z0)n.

Hence for z 6= z0,

g(z)− g(z0) =f(z)− f(z0)

z − z0− f ′(z0) =

∞∑n=2

an(z − z0)n−1.

Andg(z)− g(z0)

z − z0=

∞∑n=2

an(z − z0)n−2,

The right hand side, a holomorphic function, has a limit as z → z0 and g is differen-tiable at z0.

�

Lemma 6.1.2 (Cauchy’s Integral formula for the annulus) Let f be a holomorphicfunction in an open set containing the annulus A = {z : r1 < |z − z0| < r2} where0 < r1 < r2 <∞. Then for all z ∈ A,

f(z) =1

2πi

∫C(z0,r2)

f(ζ)

ζ − zdζ − 1

2πi

∫C(z0,r1)

f(ζ)

ζ − zdζ.

Proof We define on A,

g(ζ) =

{f(ζ)−f(z)

ζ−z , ζ 6= z

f ′(z), ζ = z.

Then g is holomorphic on the annulus. ( Take a disc D around z, which is completelycontained in A. Then f is holomorphic on D, apply the previous lemma.)

We apply the keyhole technique Lemma 5.1.1 to the function g:∫C(z0,r1)

g(ζ)dζ =

∫C(z0,r2)

g(ζ)dζ.

(c.f. Example sheet 6, Problem 4 where we take z1 = z2 and Lemma 5.1.1 in caser1 = 0) We observe further,∫

C(z0,r1)

1

ζ − zdζ = 0,

∫C(z0,r2)

1

ζ − zdζ = 2πi,

as z is outside of the circle of integrationC(z0, r1) in the first case, and inside the circleof integration D(z0, r2) in the second case.∫C(z0,r1)

g(ζ)dζ =

∫C(z0,r1)

f(ζ)− f(z)

ζ − zdζ =

∫C(z0,r1)

f(ζ)

ζ − zdζ +

∫C(z0,r1)

f(z)

ζ − zdζ

=

∫C(z0,r1)

f(ζ)

ζ − zdζ.

57

Consequently,∫C(z0,r2)

f(ζ)

ζ − zdζ =

∫C(z0,r2)

f(ζ)− f(z)

ζ − zdζ +

∫C(z0,r2)

f(z)

ζ − zdζ

=

∫C(z0,r1)

f(ζ)− f(z)

ζ − zdζ + 2πif(z)

=

∫C(z0,r1)

f(ζ)

ζ − zdζ + 2πif(z).

The conclusion follows. �

Let r1 < r < r2. If f is holomorphic on the entire disc D(z0, r2), then for z inthe annulus, f(ζ)

ζ−z is holomorphic in a disc containing D(z0, r1), its integral along thecircle |z − z0| = r1 is zero by Cauchy’s theorem:

1

2πi

∫C(z0,r1)

f(ζ)

ζ − zdζ = 0.

Theorem 6.1.3 Let r1, r2 be two real numbers with 0 ≤ r1 < r2 ≤ ∞. Suppose thatf is holomorphic in a neighbourhood of the annulus {z : r1 < |z − z0| < r2}. Thenthere exists a unique sequence of complex numbers an such that

f(z) =

∞∑n=−∞

an(z − z0)n, r1 < |z − z0| < r2.

For any r1 < r < r2, for each n,

an =1

2πi

∫C(z0,r)

f(ζ)

(ζ − z0)n+1dζ. (6.1.1)

Proof Let z ∈ {z : r1 < |z − z0| < r2}. Then by Lemma 6.1.2,

f(z) =1

2πi

∫C(z0,r2)

f(ζ)

ζ − zdζ − 1

2πi

∫C(z0,r1)

f(ζ)

ζ − zdζ.

By the same proof as in the proof for Cauchy’s theorem on Taylor formula, c.f. Theo-rem 5.3.1, we expand 1

ζ−z to obtain a power series expansion for the first term

1

2πi

∫C(z0,r2)

f(ζ)

ζ − zdζ =

∞∑n=0

an(z − z0)n

where an is as in formula (6.1.1). We work on

− 1

2πi

∫C(z0,r1)

f(ζ)

ζ − zdζ.

We observe that

1

ζ − z=

1

(ζ − z0)− (z − z0)= − 1

z − z0

1

1− ζ−z0z−z0

.

58

Since r1 = |ζ − z0| < |z − z0|, we expand 1

1− ζ−z0z−z0

as a power series and obtain:

1

ζ − z= − 1

z − z0

∞∑n=0

(ζ − z0)n

(z − z0)n= −

∞∑n=1

(ζ − z0)n−1

(z − z0)n.

By the uniform convergence of the series on {z : r1 < |z − z0|} (see remark below),we may exchange the order of taking limit and summing up and obtain

− 1

2πi

∫C(z0,r1)

f(ζ)

ζ − zdζ =

∞∑n=1

(z − z0)−n1

2πi

∫C(z0,r1)

f(ζ)(ζ − z0)n−1dζ.

Let m = −n and

am =1

2πi

∫|ζ−z0|=r

f(ζ)

(ζ − z0)m+1dζ.

Note the integration is independent of the radius of the circle, where r1 < r < r2, toconclude the theorem.

To prove the uniqueness suppose that we have another sequence of numbers suchthat

f(z) =

∞∑n=−∞

bn(z − z0)n, r1 < |z − z0| < r2.

Thenf(z)

(z − z0)k=

∞∑n=−∞

bn(z − z0)n−k.

Integrate both sides on the circle C(z0, r), with r ∈ (r1, r2), the only non-zero contri-bution comes from the term bk−1/(z − z0) which is:∫

C(z0,r)

bk−1(z − z0)−1dz = 2πibk−1.

Consequently,

bk =1

2πi

∫C(z0,r)

f(z)

(z − z0)k+1dz.

�

Remark 6.1.4 Since the annulus and its closure are contained in contained in the re-gion on which f is holomorphic, by slightly enlarging the annulus we may also taker = r1 ot r = r2 in (6.1.1).

Remark 6.1.5 Suppose that the power series function f(z) =∑∞n=0 bn(z − z0)n has

radius of convergence R > 0. Then g(z) =∑∞n=0 bn(z − z0)−n is convergent for

|z − z0| > 1R , and the convergence is uniform on any set |z − z0| ≥ 1

r for any r < R.Just note that g(z) = f( 1

z−z0 ).

Note that the radius of convergence for the positive part of the Laurent series, in Theo-rem 6.1.3, is greater or equal to r2, the negative part converges on |z − z0| > r1.

59

Example 6.1.6 Let

f(z) =1

(z − 1)(z − 3)=

1

2(

1

z − 3− 1

z − 1).

Develop the Laurent series for f on the following regions:

{|z| < 1}, {1 < |z| < 3}, {|z| > 3}.

We observe that z0 = 0 and

1

z − 1=

−∑∞n=0 z

n, |z| < 1∑∞n=1 z

−n, |z| > 1.

1

z − 3=

= − 1

3 ·1

1− z3= −

∑∞n=0 3−n−1zn, |z| < 3

1z ·

11− 3

z

=∑∞n=1 3n−1z−n, |z| > 3.

Hence on {|z| < 1}, where f is holomorphic,

f(z) =1

2

(−∞∑n=0

3−n−1zn +

∞∑n=0

zn

)=

1

2

∞∑n=0

(1− 3−n−1

)zn.

On {1 < |z| < 3},

f(z) =1

2

(−∞∑n=0

3−n−1zn −∞∑n=1

z−n

)= −1

2

∞∑n=1

z−n − 1

2

∞∑n=0

3−n−1zn.

On |z| > 3,

f(z) =1

2

( ∞∑n=1

3n−1z−n −∞∑n=1

z−n

)=

1

2

−1∑n=−∞

(3−n−1 − 1)zn.

Corollary 6.1.7 (Cauchy’s Inequality) The coefficients in the Lauren series expan-sion f(z) =

∑∞n=−∞ an(z − z0)n satisfies:

|ak| ≤ sup|z−z0|=r

|f(z)| 1

rk

for any r ∈ (r1, r2).

Proof Just use the formula

|ak| =

∣∣∣∣∣ 1

2πi

∫C(z0,r)

f(ζ)

(ζ − z0)k+1dζ

∣∣∣∣∣ ≤ 1

2πsup

|z−z0|=r|f(z)|

(1

r

)k+1

L

where L = 2πr is the length of the circle. The conclusion follows. �

60

6.2 Classification of Isolated Singularities (Lecture 19)Denote D′(z0, r) = {z : 0 < |z − z0| < r} the deleted disc. Such a disc is a deletedneighbourhood of z0. A full disc D(z0, r) = {z : |z − z0| < r} is a neighbourhood ofz0. Both are assumed to be fully contained in the open set on which a function underdiscussion is defined.

In this section we discuss isolated singularities. A point z0 is an isolated singularityfor f , if f is defined and holomorphic on a deleted disc D′(z0, R), in which case f hasa Laurent series at z0:

f(z) =

∞∑n=−∞

an(z − z0)n, z ∈ D′(z0, R).

Definition 6.2.1 1. If ak 6= 0 for an infinite number of negative integers, we say fhas an essential singularity at z0.

2. If −n is the least integer for which a−n 6= 0, and n is positive, we say f has apole at z0 of order n.

3. If ak = 0 for all negative k, we say z0 is a removable singularity.

We have not assumed that f is defined at z0. If an = 0 for all negative integers,we extend f to f by defining f(z0) = a0. Then the extended function is holomorphic,hence the terminology f has a removable singularity at z0.

The equivalence of (1) with (4) in the theorem below is called Riemann’s theoremon removable singularity.

Theorem 6.2.1 Let f be a holomorphic function on the deleted disc D′(z0, R). Thefollowing statements are equivalent.

1. ** f has a removable singularity at z0;

2. f can be analytically continued to the whole disc D(z0, R);

3. limz→z0 f(z) exists in C (i.e. the limit is a complex number);

4. ** f is bounded in a deleted neighbourhood of z0. (i.e. there exists δ such thatsup0<|z−z0|<δ |f(z)| <∞).

Proof (1)=⇒ (2). Suppose f(z) =∑∞n=0 an(z − z0)n. We define f(z0) = a0 and f

extends analytically to the whole disc.(2) =⇒ (3). If f is the analytic continuation, then limz→z0 f(z) = f(z0), a finite

number;(3)=⇒ (4) If limz→z0 f(z) = c, then it is clear f(z) is bounded in a neighbourhood

of z0. (Take ε = 1, there exists R > δ > 0 such that if |z − z0| < δ, |f(z) − c| < 1and so |f(z)| ≤ |c|+ 1.)

(4)=⇒ (1) We assume that sup|z−z0|≤δ0 |f(z)| ≤ M for some positive numbersδ0 and M . By Cauchy’s inequality, on D(z0, δ) where δ < δ0, the coefficients is theLaurent series expansion of f satisfy:

|a−k| ≤M1

δ−k= Mδk.

We take δ → 0. If k ≥ 1, a−k = 0. Thus f has removable singularity. �

61

6.2.1 PolesRecall that f has a pole at z0 if it has Laurent expansion at z0 of the form below, wheren is a natural number and a−n 6= 0:

f(z) =a−n

(z − z0)n+

a−n+1

(z − z0)n−1+ . . .

a−1

(z − z0)+G(z)

where G is a holomorphic function. The order of f at z0, denoted by ord(f ; z0), is n.

Definition 6.2.2 If ord(f ; z0) = −1, the pole z0 is said to be simple.

Remark 6.2.2 If f has a pole of order n ∈ N at z0, then for any k > n

limz→z0

(z−z0)kf(z) = limz→z0

(a−n

(z − z0)k

(z − z0)n+ a−n+1

(z − z0)k

(z − z0)n−1+ . . . a−1

(z − z0)k

(z − z0)1

)= 0.

If k = n, limz→z0(z − z0)kf(z) = a−n, which does not vanish by the assumption. Ifk < n, limz→z0(z − z0)kf(z) =∞. Thus,

ord(f ; z0) = inf{n : limz→z0

(z − z0)nf(z) exists and is finite}.

If n > ord(f ; z0) then limz→z0(z − z0)nf(z) = 0.

Remark. If we wish to treat zeros and poles together, without discrimination, it isadvantage to define the order of a pole to be a negative integer. I am happy with eitherdefinitions, as far as in the same line of computation, we are consistent.

Definition 6.2.3 If {zk} is a sequence of complex numbers, we say limk→∞ zk = ∞if limk→∞

1zk

= 0.

Theorem 6.2.3 Let f be a holomorphic function on the deleted disc D′(z0, R). Thefollowing statements are equivalent.

1. f has a pole at z0.

2. limz→z0 f(z) =∞ (i.e. limz→z0 |f(z)| =∞).

Proof (1) =⇒ (2). If f has a pole of order n at z0, by the definition in a discD′(z0, R),

f(z) = (z − z0)−n∞∑

k=−n

ak(z − z0)k+n = (z − z0)−nh(z).

where h(z) =∑∞k=−n ak(z−z0)k+n =

∑∞j=0 aj−n(z−z0)j has an analytic extension

to D(z0, R). Since a−n 6= 0, it is routine to prove h(z0) 6= 0 and hence does notvanish in a small disc around z0, thus limz→z0

1f(z) = limz→z0

(z−z0)n

h(z) = 0, andlimz→z0 f(z) =∞.

(2) =⇒(1). Since limz→z0 f(z) = ∞, f(z) does not vanish in a sufficiently smalldisc D(z0, δ), where we define g(z) = 1

f(z) . Then limz→z0 g(z) = 0 and by Rie-mann’s theorem on removable singularity (Theorem 6.2.1), g extends to a holomorphicfunction on D(z0, δ) with g(z0) = 0 and g is not identically zero. Then there exists apositive integer n and a holomorphic function H (see Lemma 5.8.1) with H(z0) 6= 0

62

such that g(z) = (z − z0)nH(z). Let h = 1H . Then h has a Taylor expansion

h(z) =∑∞z=0 bn(z − z0)n and h(z0) 6= 0. Consequently f has a Laurent series

expansion

f(z) =h(z)

(z − z0)n= (z − z0)−n(b0 + b1(z − z0) + . . . ),

and f has a pole at z0. �

Remark 6.2.4 A holomorphic function f on D′(z0, R) has a pole of order n at z0 ifand only if there exists a holomorphic function on a disc D(z0, δ) with h(z0) 6= 0 s.t.

f(z) = (z − z0)−nh(z), z ∈ D′(z0, δ),

i.e. f(z)(z − z0)n is holomorphic and does not vanish at z0.

Definition 6.2.4 If f has a pole at z0, the number a−1 is the residue of f at the pole z0

and is denoted by Res(f ; z0).

Corollary 6.2.5 Let f be a holomorphic function on D′(z0, R) having a pole at z0,then

Res(f ; z0) =1

2πi

∫C

f(ζ)dζ,

for any circle centred at z0 with radius r < R.

If the order of the pole z0 is one then

Res(f ; z0) = limz→z0

(z − z0)f(z).

Theorem 6.2.6 If f has a pole of order n at z0, then

Res(f ; z0) = limz→z0

1

(n− 1)!(d

dz)n−1(z − z0)nf(z).

Proof Since

f(z) =

∞∑k=−n

ak(z − z0)k, z ∈ D′(z0, R),

(z − z0)nf(z) = a−n + a−n+1(z − z0) + · · ·+ a−1(z − z0)n−1 +G(z)(z − z0)n

where G is a holomorphic function, the positive part of the Laurent expansion.

(d

dz)n−1(z − z0)nf(z) = (n− 1)!a−1 + (

d

dz)n−1 (G(z)(z − z0)n) .

The last term vanishes as z → z0. �

63

6.2.2 Essential SingularitiesTheorem 6.2.7 (Casorati-Weierstrass) Let f be a holomorphic function on the deleteddisc D′(z0, R). The following statements are equivalent.

1. **f has an essential singularity at z0;

2. for all 0 ≤ r ≤ R, the image f(D′(z0, r)) is dense in C.

3. ** limz→z0 f(z) does not exist on the extended plane C∗.

Proof (2) =⇒(3): (3) follows trivially from (2).(3) =⇒(1). If (3) holds, z0 is not a removable singularity nor a pole for f . Hence

(1) holds.(1) =⇒(2). Let z0 be an essential singularity of f . Suppose (2) fails. There exists

r > 0 such that the image of D′(z0, r) by f is not dense in C. Then there exists a pointw ∈ C and δ > 0 such that |f(z)− w| > δ for all z ∈ D′(z0, r). Define

g(z) =1

f(z)− w.

Then g is holomorphic and bounded by 1δ on D′(z0, r). By Theorem 6.2.1, z0 is a

removable singularity for g. If g(z0) 6= 0 then

limz→z0

f(z) = w + limz→z0

1

g(z)

exists, then z0 would be a removable singularity for f ! If, otherwise g(z0) = 0, thenlimz→z0 f(z) =∞ and z0 is a pole for f ! Both scenario lead to contradictions and so(2) must holds. �

6.3 Meromorphic FunctionDefinition 6.3.1 A meromorphic function on an open set U is a function which arecomplex differentiable everywhere with the exception of isolated singularities that arepoles.

Remark 6.3.1 A meromorphic function has only a finite number of poles on a boundedclosed subset of U .

Proof Indeed, if there is an infinite number of singularities in a closed subset A, ithas an accumulation point z0 in A which cannot be a singularity by the assumption.However f cannot be differentiable at z0 either, for otherwise it would be bounded in aneighbourhood D of z0 on one hand, and approaches infinity near the poles within Don the other hand. �

6.3.1 SupplementaryRemark 6.3.2 The set of meromorphic functions, with function addition, multiplica-tion, division, is a group.

64

Proof It is clear that the addition, subtraction, and multiplication of two meromorphicfunctions are meromorphic. Let us consider the quotient of two holomorphic functionsf and g. Suppose that f or g has an isolated zero or an isolated pole at z0. Then nearz0,

f(z) = (z − z0)mh(z), g(z) = (z − z0)nk(z)

where h, k are holomorphic functions. Then

f(z)

g(z)= (z − z0)m−n

h(z)

k(z)

is holomorphic in a deleted disc about z0. �

65

Chapter 7

Winding Numbers and theResidue Theorem

The function f(z) = 1z does not have a primitive in an open set of C containing a

disc about the origin, for its integral along the boundary of the disc is 2πi 6= 0. If weintegrate the 1

z along a square centred at the origin, the integral is also 2πi (on eachside of the line we may choose a brach of the logarithm). If we integrate it along thecircle concatenated with itself n− 1 times we see 2πni.

Lemma 7.0.1 If a piecewise smooth closed curve γ does not pass through a point z0

then ∫γ

dz

z − z0

is a multiple of 2πi.

Proof Let z : [a, b]→ C be a parameterization of γ, then∫γ

dz

z − z0=

∫ b

a

z′(s)

z(s)− z0ds.

Define

g(t) =

∫ t

a

z′(s)

z(s)− z0ds, t ∈ [a, b].

It is sufficient to prove that e−g(b) = 1, from which g(t) = 2kπi where k ∈ Z.An obvious candidate for g(t) is ln z(t)−z0

z(a)−z0 . Define

F (t) = e−g(t)z(t)− z0

z(a)− z0.

Then for t ∈ [a, b] where z is C1, we have

F ′(t) = e−g(t)(− z′(t)

z(t)− z0)z(t)− z0

z(a)− z0+ e−g(t)

z′(t)

z(a)− z0= 0.

Since F is continuous in t, and piecewise a constant , then F must be a constant on[a, b]. Since g(a) = 0, F (a) = 1. Note z(b) = z(a),

F (b) = e−g(b)z(b)− z0

z(a)− z0= e−g(b) = 1.

66

�

7.1 The Index of a Closed Curve (Lecture 22)Definition 7.1.1 If a piecewise smooth closed curve γ does not pass through a pointz0 then the index of γ with respect to z0 is:

ind(γ, z0) =1

2πi

∫γ

dz

z − z0.

It is also called the winding number or rotation number of the curve γ about z0.

We recall the following facts. Any curve z : [a, b] → C can be reparameterized to[0, 1]. z(t) = t(b − a) + a, t ∈ [0, 1]. If γ has a parameterization z : [0, 1] → C then−γ is the curve defined by z(t) = z(1 − t). Let i = 1, 2 and γi be two curves withrespectively the parameterizations zi : [0, 1]→ C with z1(1) = z2(0). Then we defineγ be the concatenated curve given by

z(t) =

{z1(2t), 0 ≤ t ≤ 1

2z2(1− 2t), 1

2 ≤ t ≤ 1.

This curve is denoted by γ1 ∪ γ2, otherwise denoted by γ1 + γ2.

Remark 7.1.1 1. For any z0 not passing through the curve,

ind(−γ, z0) = − ind(γ, z0).

2. If γ1 and γ2 are two smooth closed curves with the same initial points, then forz0 not passing through γ1 nor γ2,

ind(γ1 ∪ γ2, z0) = ind(γ1, z0) + ind(γ2, z0).

Definition 7.1.2 A curve γ : [a, b] → C is simple if it does not have self intersection.That is: γ(t) 6= γ(s) unless s = t or a = b.

By the boundary of a star region we mean the curve which travels along the boundaryonce, with positive orientation. This is a simple curve.

Example 7.1.2 Let R denote the square centred at the origin. By standard theorem,∫R

1zdz = 2πi. However we compute the integral side by side to see the ‘winding’.

We work with the top horizontal side, which is contained in a slit domain where ithas a primitive. Evaluate the difference of the end points, we find the change in theargument: π2 . By symmetry this is π/2 for the other sides. All together we obtain:

ind(R, 0) =1

2πi

∫R

1

zdz = 1.

The curve in the example below is not a simple curve.

Example 7.1.3 The curve z(t) = z0 + eint, 0 ≤ t ≤ 2π is the circle concatenatedwith itself n− 1 times, it is not simple.

ind(γ, z0) =1

2πi

∫γ

dz

z − z0= n.

67

Let γ be a smooth curve with parameterization z : [a, b]→ C. Then its complementis an open set which we denote by U . Any (connected) component of an open set is anopen set and there are only a countable number of components. Indeed, if z belongs toan open connected component V of U , let D(z, r) be a disc in U , then V ∪D(z, r) isconnected. Hence V is open. Also, the set {q1 + q2i ∈ U : q1, q2 ∈ Q} is countable,each connected component contains a disc and hence one of these points.

Let M = supt∈[a,b] |f(z(t))|, which is finite. Then for any R > M , the set{z : z| > R} belongs to the complement of γ in C. Hence γ has only one unboundedcomponent.

Denote by {γ} the set of points on the curve γ.

Theorem 7.1.4 Let γ be a closed piecewise smooth curve and U = C \ {γ}. Thenind(γ, z) is a constant of z when restricted to a component of U . If z belongs to theunbounded component of U then ind(γ, z) = 0.

Proof We prove that z 7→ ind(γ, z) is a continuous from U to C. Since ind(γ, ·) isinteger valued, it has to be a constant on each connected component of U .

Let V be a connected component of U and z0 ∈ V and R the distance from z0 toγ. Let δ = R

3 , then V contains the D(z0, δ). For any z with |z − z0| ≤ δ,

ind(γ, z)− n(γ, z0) =1

2πi

∫γ

(1

ζ − z− 1

ζ − z0

)dζ.

Then

| ind(γ, z)− n(γ, z0)| = 1

2π

∣∣∣∣∫γ

z − z0

(ζ − z)(ζ − z0)dζ

∣∣∣∣ ≤ l(γ)

2π

1

δ2|z − z0|,

as |ζ−z0| ≥ δ and |ζ−z0| ≥ δ for ζ ∈ γ and z ∈ D(z0, r). This proves the continuity.Let z0 be a point in the unbounded component of U , which contains any z with

|z| > supw∈γ |f(w)|. For such z,

ind(γ, z0) = ind(γ, z) =1

2πi

∫γ

1

ζ − zdζ.

But by taking z far away, ind(γ, z0) can be take to be arbitrarily close to 0 and istherefore zero.

�

Example 7.1.5 If U is a simply connected domain with γ its smooth boundary withpositive orientation, then

ind(γ, z0) =

{1, if z0 ∈ U0, If z0 ∈ C \ U .

Definition 7.1.3 A closed piecewise smooth curve γ in an open set U is said to behomologous to zero in U , if ind(γ, z0) = 0 for all z0 ∈ C \ U . This is denoted byγ ≈ 0.

Example 7.1.6 If γ is a closed piecewise smooth curve in U , homotopic to 0, thenγ ≈ 0.

68

The curve γ can be deformed continuously to a curve γ in a disc D, the discis contained entirely in U . But 1

z−z0 is a holomorphic function for any z0 6∈ D,and ind(γ, z0) = 0 by Cauchy’s theorem. Since the index of a curve is an integer,ind(γ,w) = ind(γ, w), and ind(γ, z) = 0 for z ∈ C \ U . This is not a proof, as theintermediate deformation from γ to the constant curve may not be piecewise C1.

Theorem 7.1.7 (Cauchy’s integral formula) Let f be a holomorphic function in anopen set U . If γ1, . . . , γm are closed piecewise C1 curves s.t.

m∑k=1

ind(γk, w) = 0, ∀w ∈ C \ U,

then for any z0 ∈ U \ ∪∞k=1{γk},

f(z0)

m∑k=1

ind(γk, z0) =

m∑k=1

1

2πi

∫γk

f(z)

z − z0dz.

The proof is given in the supplementary section.

Theorem 7.1.8 (Cauchy’s Theorem) Let f be a holomorphic function in an open setU . If γ1, . . . , γm are closed piecewise C1 curves s.t.

m∑k=1

ind(γk, w) = 0, ∀ w ∈ C \ U,

thenm∑k=1

∫γk

f(z)dz = 0.

Proof In Cauchy’s Integral formula substitute f(z)(z − z0) for f(z) at a point z0 ∈U \ ∪∞k=1{γk}. �

Example 7.1.9 1. If γ is homologous to zero,∫γf(z)dz = 0.

2. If γ is a closed curve in a simply connected region then∫γf(z)dz = 0.

7.1.1 SupplementaryLemma 7.1.10 For f defined and continuous on a piecewise C1 curve γ, the function∫γf(w)−f(z)

w−z dw is holomorphic on C \ {γ}. Furthermore for each integer m ≥ 1,define for z 6∈ {γ},

Fm(z) =

∫γ

f(w)

(w − z)mdw

then each Fm is holomorphic on C \ {γ} and F ′m(z) = mFm+1(z).

Proof This can be proved by the definition of complex differentiation, and by inductionon m. �

We supply one proof for Cauchy’s integral formula.

69

Theorem 7.1.11 (Cauchy’s integral formula) Let f be a holomorphic function in anopen set U . If γ1, . . . , γm are closed piecewise C1 curves s.t.

m∑k=1

ind(γk, w) = 0, ∀w ∈ C \ U,

then for any z0 ∈ U \ ∪∞k=1{γk},

f(z0)

m∑k=1

ind(γk, z0) =

m∑k=1

1

2πi

∫γk

f(z)

z − z0dz.

Proof We prove the case where there is only one curve γ, the proof for m ≥ 2 isanalogous. We wish to prove that for z0 ∈ U \ {γ},∫

γ

f(w)− f(z0)

w − z0dw = 0.

If so, ∫γ

f(w)

w − z0dw =

∫γ

f(z0)

w − z0dw = f(z0) ind(γ, z0)2πi.

Define g : C \ {γ} → C by

g(z) =

∫γ

f(w)− f(z)

w − zdw.

By Lemma 7.1.10, g is holomorphic on C\{γ}. We construct an analytic continuationof g to the whole plane and prove the resulting entire function is a constant.

LetH = {z ∈ C \ {γ} : ind(γ, z) = 0}.

If z ∈ H ,∫γ

f(z)

w − zdw = f(z) ind(γ, z)2πi = 0, g(z) =

∫γ

f(w)

w − zdw. (7.1.1)

Since the index is integer valued and ind(γ, z) is continuous in z, thenH is an open set(as H = {z ∈ C : ind(γ, z) < 1

2}). Also C = U ∪H following from the assumptionthat γ is homologous to zero (and so C \ U is a subset of H).

It remains to define g on {γ}. We define φ : U × U → C by

φ(z, w) =

{f(w)−f(z)

w−z , z 6= w

f ′(z), z = w.

Then φ is continuous. For each w ∈ C, φ(·, w) is holomorphic (Lemma 6.1.1). Define

g(z) =

{ ∫γφ(z, w)dw, z ∈ U∫

γf(w)w−z dw, z ∈ H .

If z ∈ U ∩H , z 6∈ {γ}, the two formulas agree:∫γ

φ(z, w)dw =

∫γ

f(w)− f(z)

w − zdw =

∫γ

f(w)

w − zdw,

70

by (7.1.1), and g is well defined. Furthermore g is an entire function.We observeH contains the unbounded component of C\{γ} and thus a neighbour-

hood of∞ in the extended complex plane. Since f is bounded on γ and limz→∞1

z−w =0 uniformly on γ,

limz→∞

g(z) = limz→∞

∫γ

f(w)

w − zdw = 0. (7.1.2)

By the extension to Liouville’s theorem, Example 5.4.5, g is a constant on C. Sincelimz→∞ g(z) = 0 then g is identically zero. The proof is complete. �

7.2 The Residue Theorem (Lecture 23)Recall that a meromorphic function has only a finite number of poles on a boundedsubset ofU . Also ifw belongs the unbounded component of C\{γ}where γ is a closedpiecewise C1 smooth curve, ind(γ,w) = 0. In particular the set {w : ind(γ;w) 6= 0}is a bounded set.

Theorem 7.2.1 (The Residue Theorem) Let f be a meromorphic function in a regionU with a finite number of poles {z1, . . . , zm}. If γ is a closed piecewise C1 smoothcurve in U , not passing through any of the poles {z1, . . . , zm}, and γ ≈ 0 in U . Then

1

2πi

∫γ

f(z)dz =

m∑k=1

ind(γ, zk) Res(f ; zk).

Proof Let V = U \ {z1, . . . , zm} then f is holomorphic on V . We choose positivenumbers δk such that the discs D(zk, δ) are contained in U and no two discs intersect.Define

γk(t) = zk + δke−it·ind(γ,zk), 0 ≤ t ≤ 2π.

Then they are all homologous to 0 in U and

ind(γk, zj) = − ind(γ, zk)δkj .

We thus have

ind(γ,w) +

m∑k=1

ind(γk, w) = 0, ∀ w ∈ C \ V.

By Cauchy’s theorem (Theorem 7.1.8),∫γ

f(z)dz +

m∑k=1

∫γk

f(z)dz = 0.

Now by the Laurent’s expansion for f in the deleted disc D′(zk, δk),∫γk

f(z)dz =

∫γk

Res(f ; zk)

z − zkdz = 2πiRes(f ; zk) ind(γk, zk)

= −2πRes(f ; zk) ind(γ, zk).

Since∫γf(z)dz = −

∑mk=1

∫γkf(z)dz, this completes the proof. �

71

7.3 Compute Real IntegralsRule of Thumb. Let P and Q be polynomial of degree n and m respectively withm ≥ n+ 2. We wish to compute real integrals of the form∫ ∞

−∞

P (x)

Q(x)dx.

Let us take the contour curve γR = [−R,R] ∪ C+R , where C+

R is the upper half of thecircle of radius R centred at 0. Then on one hand, we may apply the Residue theorem,to obtain for sufficiently large R,∫

γR

P (z)

Q(z)dz = 2πi

∑ζ

Res(P

Q; ζ).

The summation is over poles inside the contour. On the other hand, as R→∞.∣∣∣∣∣∫C+R

P (z)

Q(z)dz

∣∣∣∣∣ ≤ C

R→ 0.

This means ∫ ∞−∞

P (x)

Q(x)dx = 2πi

∑ζ

Res(P

Q; ζ).

Example 7.3.1 Suppose that p, q are holomorphic functions in a neighbourhood of z0

and f = pq has a simple pole at z0. If p(z0) 6= 0, then

Res(f ; z0) =p(z0)

q′(z0). (7.3.1)

In fact,

Res(f ; z0) = limz→z0

(z − z0)p(z)

q(z)= p(z0) lim

z→z0

z − z0

q(z)=

p(z0)

q′(z0),

This follows from the Taylor expansion:

q(z) = q(z0) + q′(z0)(z − z0) + (z − z0)2∞∑k=2

ak(z − z0)k−2.

Example 7.3.2 Prove that ∫ ∞−∞

dx

x6 + 1=

2π

3.

Let us consider thef(z) =

1

z6 + 1.

On the region bounded by the line segment [−R,R] and the semi-circle C+, f hasthe following singularities:

c1 = eiπ6 , c2 = e

3πi6 = i, c3 = e

5πi6 .

72

The residues at these points are, using formula (7.3.1):

Res(f ; c1) =1

6e−

5πi6 , Res(f ; c2) = −1

6i, Res(f ; c3) =

1

6e−iπ6 .

By the residue theorem,∫γ

f(z)dz =

∫ R

−R

dx

x6 + 1+

∫ 2π

0

iReit

R6ei6t + 1dt = (2πi)

3∑i=1

Res(f ; ci).

But (2πi)∑3i=1 Res(f ; ci) = (2πi)(− 2i

6 ) = 2π3 and

limR→∞

∫ 2π

0

iReit

R6ei6t + 1dt = 0.

Consequently,∫∞−∞

dxx6+1 = 2π

3 .

73

Chapter 8

Fundamental Theorems

Any non-zero number z has a logarithm, so elog z = z. As z varies, log z would definean inverse function to ez , which we hope to be holomorphic. For z ∈ C \ {(−∞, 0)},we may write z = reiθ where θ ∈ (−π, π) and define the principal branch of thelogarithm, on the slit plane C \ {(−∞, 0)}, by

log z = log r + iθ.

If f(z) is a never vanishing function in a region U , for which domains could wechoose the logarithm of f(z) in a way so that the resulting function in holomorphic?

Theorem 8.0.1 Let U be a star region (or more generally a simply connected region).Suppose that f(z) is holomorphic and never vanishing in U . Then there exists a holo-morphic function h : U → C such that eh(z) = f(z).

Proof Since f(z) 6= 0 for any z ∈ U , f ′

f is holomorphic in U and hence has aprimitive, by theorem 4.4.3). (More generally, a holomorphic function in a simplyconnected region has a primitive).

Let h be a primitive of f ′

f , i.e. a holomorphic function such that h′ = f ′

f . Setg(z) = e−h(z)f(z), then g′(z) = −h′(z)e−h(z)f(z) + e−h(z)f ′(z) = 0. So g is aconstant, say k, and k = e−h(z0)f(z0) for a given point z0 ∈ U . Choose a primitive hof f

′

f such that eh(z0) = f(z0) to conclude. �

8.1 The Argument Principle (Lecture 24)The logarithm of a holomorphic never vanishing function f is mutli-valued, log f(z) =

log |f(z)|+ i arg(z), its derivative is single-valued (log f(z))′ = f ′(z)f(z) .

If a holomorphic function f has a zero of orderm at z0, then f(z) = (z−z0)mg(z)where g is holomorphic and g(z0) 6= 0,

f ′(z)

f(z)=

m

z − z0+g′(z)

g(z).

The function g is holomorphic near z0. Suppose f is meromorphic and has a pole oforder m at z0. Then f(z) = (z − z0)−mg(z) where g is holomorphic and g(z0) 6= 0.

74

Also,f ′(z)

f(z)= − m

z − z0+g′(z)

g(z).

Theorem 8.1.1 Let f be a meromorphic function in a region U with poles p1, . . . , pnand zeros z1, . . . , zm, counted according to multiplicity. Let γ be a closed piecewisesmooth curve γ in U with γ ≈ 0, and not passing through p1, . . . , pn, z1, . . . , zm. Then

1

2πi

∫γ

f ′(z)

f(z)dz =

m∑j=1

ind(γ, zj)−n∑i=1

ind(γ, pi).

Proof By repeated application of the discussion above,

∫γ

f ′(z)

f(z)dz =

∫γ

n∑k=1

1

z − zk−

m∑j=1

1

z − pj+g′(z)

g(z)

dz

where g is holomorphic, g′/g is holomorphic. By Cauchy’s theorem, since γ ≈ 0,∫γ

g′(z)

g(z)dz = 0.

The rest follows from the definition of the index. �

Apply the above theorem to a holomorphic function we obtain the following:

Theorem 8.1.2 Let U be a simply connected region with contour γ. Let f be a mero-morphic function in an open set containing the closure U with poles p1, . . . , pn and ze-ros z1, . . . , zm, counted according to their multiplicity, none of which passing throughthe curve γ. Then ∫

γ

f ′(z)

f(z)dz = Z(f)− P (f)

where Z(f) and P (f) are respectively the number of zeros and poles of f in U .

Interpretation of the theorem. Suppose f is holomorphic. If γ a closed piecewiseC1 curve, so is the pushed forward curve f ◦ γ, and it is easy to see (c.f. exercise 3,sheet 4),

1

2πi

∫γ

f ′(z)

f(z)dz =

1

2πi

∫f◦γ

1

zdz = ind(f ◦ γ, 0).

8.2 Rouche’s TheoremRouche’s theorem states that the number of zeros minus the numbers of poles, Z(f)−P (f), of a function f is unchanged if we modify f slightly. If on a closed curve, thedistance of the change g(z) to the origin is smaller than the distance of f(z) to theorigin, then f and f + g have the same number of zeros minus poles inside the curve.We state a simple version of the theorem.

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Theorem 8.2.1 Suppose that f, g are holomorphic in an open set containing a closeddisc D(z0, R). Let Z(f) and Z(g) denote respectively the number of zeros of f and ginside the circle γ = C(z0, R), counted according to their multiplicity. Suppose that

|g(z)| < |f(z)|, ∀z ∈ {γ}. (8.2.1)

Then Z(f + g) = Z(f).

Proof Let h(z) = f(z)+g(z)f(z) = 1 + g(z)

f(z) . By (8.2.1), f has no zero on {γ}, f + g hasno zero on {γ} and h is a holomorphic around γ. Indeed, On {γ}, |f(z)| > |g(z)| > 0and |f + g| ≥ |f | − |g| > 0. Furthermore,

h′

h=

(f + g)′

f + g+f ′

f

Integrate the above identity and apply Theorem 8.1.2 to f + g and f to obtain:∫γ

h′

h= Z(f + g)− Z(f).

But1

2πi

∫γ

h′(z)

h(z)dz =

1

2πi

∫h◦γ

1

zdz.

But there exists δ < 1 such that |h(z) − 1| < δ for any z ∈ {γ}. This means {h(γ)}is contained in the disc D(1, δ) which does not contain 0 and ind(h ◦ γ, 0) = 0. Thisconcludes Z(f + g)− Z(f) = 0. �

Example 8.2.2 This gives another proof of the fundamental theorem of algebra. Ifpn(z) is a polynomial of degree n. Write pn(z) = anz

n + Qn−1(z). Then for Rsufficiently large, |Q(z)| ≤ |an||zn|, so pn(z) has the same number of zero’s as zn

inside |z| ≤ R, while zn = 0 has zero or order n at 0.

8.2.1 SupplementaryExample 8.2.3 If f is holomorphic in a neighbourhood of the closed disc D = {z : |z| ≤ 1}and |f(z)| < 1 for |z| = 1. There there exists exactly one solution of f(z) = z inD = {z : |z| < 1}.

This is an easy consequence of Rouche’s theorem: treat f(z) as a perturbation toz, |f(z)| < |z| on the circle {z : |z| = 1}, hence z − f(z) has one solutions in D, justas g(z) = z.

8.3 The Open mapping Theorem (Lecture 25)Recall that f has a zero of order m ≥ 1 at z0 if in a neighbourhood of z0, f(z) =∑∞k=m ak(z−z0)k. Equivalently, f(z0) = 0, . . . , f (m−1)(z0) = 0, and f (m)(z0) 6= 0.

In particular f is not a constant.

Theorem 8.3.1 Let f be a non-constant holomorphic function in D(z0, R). Supposethat f has a zero of order m ≥ 1 at z0. If ε > 0 is sufficiently small, then thereexists a corresponding positive number δ such that for w ∈ D′(f(z0), δ) the equation

76

f(z) = w has exactly m solutions in z ∈ D′(z0, ε). In particular D(f(z0), ε) iscontained in the image of D(z0, ε) by f , i.e. D(f(z0), ε) ⊂ f(D(z0, ε)). Each zero off(z)− w, for w ∈ D′(f(z0), δ), is a simple zero.

Proof For w ∈ C define Fw(z) = f(z)− w. Set w0 = f(z0). Since z0 is an isolatedzero of Fw0

(z) = f(z) − w0, there exists a number ε > 0 such that 3ε < R and onD′(z0, 3ε) := {z : 0 < |z − z0| < 3ε} the following statements hold:

1. Fw0(z) ≡ f(z)− w0 6= 0,

2. F ′w0(z) ≡ f ′(z) 6= 0.

(If m = 1, then F ′w0(z0) = f ′(z0) 6= 0. Since F ′w0

is continuous, it does not vanish ina neighbourhood of z0.If m ≥ 1, then F ′w0

(z0) = 0 in which case z0 is an isolated zero for the holomorphicfunction F ′w0

.)We observe that Fw(z) = f(z) − w0 − (w − w0) = Fw0

(z) − (w − w0). As|f(z)−w0| is continuous and non-zero on the circle, δ = inf |z−z0|=ε |f(z)−w0| > 0,by the compactness of the circle. If |w−w0| < δ, so in particular |w−w0| < |Fw0

(z)|on the circle, we apply Rouche’s theorem to see Fw and Fw0 have the same number ofzeros’s inside the disc {z : |z − z0| < ε}.

In particular each w ∈ {w : |w − w0| < δ} has a pre-image in {z : |z − z0| < ε}.For w with 0 < |w − w0| < δ, let z be a-pre-image of w, then F ′w(z) 6= 0 and z mustbe a simple root of Fw. �

A map f is open if it takes an open set to an open set. Theorem 8.3.1 leads, trivially,to the following corollary.

Theorem 8.3.2 (The Open Mapping Theorem) A non-constant holomorphic map ina region is an open map.

Remark 8.3.3 Suppose that f : U → C is holomorphic. Then f is injective in aneighbourhood of z0 ∈ U if and only if f ′(z0) 6= 0. (This is a local statement, only.The holomorphic function f : C \ {0} → C given by f(z) = z2 is not globallyinjective, f ′(z) 6= 0.) In particular if f is one to one then f ′ never vanishes andf−1 : f(U)→ U is holomorphic (by Theorem 1.8.3, The Inverse Function Theorem);consequently f preserves angle at each point.

Lemma 8.3.4 Suppose f is a holomorphic and non-constant in a region U . Then forany z0 ∈ U and any R > 0 such that D(z0, R) ⊂ U , there exists z ∈ D(z0, R) suchthat |f(z)| > |f(z0)|.

Proof By the open mapping theorem, f(D(z0, R)) contains a disc D(f(z0), δ), in thelatter disc there existsw such that |w| ≥ |f(z0)|. (e.g. ifw0 = r0e

iθ0 takew = r0eiθ0+

δ2eiθ0 , |w| = |w0|+ δ

2 ). The pre-image of w in D(z0, R) satisfies |f(z)| > |f(z0)|. �

Theorem 8.3.5 (Maximal Modulus Principle) If f is a non-constant holomorphic func-tion in a region U , then |f | cannot attain a local maximum in U .

Proof If f attains its maximum at z0 ∈ U , it is a maximal in a disc D(z0, r), where ris small so that D(z0, r) ⊂ U . This is impossible from the previous lemma. �

77

Corollary 8.3.6 Let U be a region with compact closure U . Let f : U → C becontinuous and holomorphic on U . Then supz∈U |f(z)| is attained on the boundary ofU .

Proof Since f is continuous on the compact set U , it attains its maximum in U , whichcannot be attained in the interior. �

8.3.1 SupplementaryWhy the argument for proving Maximal Modulus Principle might fail for the infimum,inf |f(z)|? Think of the case w0 = 0! In fact if the function f never vanishes, wewould have no problem, just apply the Maximal Modulus Principle to 1

f .

Corollary 8.3.7 If f is a holomorphic function on a disc D(z0, R) and f(z0) 6= 0.Then for all 0 < r < R there exists z such that |f(z)| < |f(z0)|.

Proof Since f(a) 6= 0, f does not vanish in a disc D(z0, r0), r0 < R then 1f(z)

is holomorphic on D(z0, r0). There exists a point z in the disc D(z0, r0) such that∣∣∣ 1f(z)

∣∣∣ > ∣∣∣ 1f(z0)

∣∣∣. �

Example 8.3.8 Let f be a not constant holomorphic function in a neighbourhood ofthe closed unit disk D := {z : |z| ≤ 1}. Assume that |f(z)| is constant on ∂D :={z : |z| = 1}. Then f must attain at least one zero in D.

Suppose |f(z)| = 0 on the unit circle, then |f(z)| = 0 inside the disc, by themaximal principle, and the assertion holds trivially. Suppose |f(z)| = a > 0 onthe unit circle then |f(z)| is non-zero in a neighbourhood of the unit circle. Assumethat f does not have a zero in D. Then 1

f is holomorphic in D := {z : |z| < 1}and continuous on D. Then the supremum and infimum of |f(z)| are attained on theboundary: supz∈D |f(z)| = infz∈D |f(z)| = a. Hence |f(z)| is a constant on thewhole disc. By the Maximal modulus theorem this implies that f is constant.

The following is an alternative proof for the maximal modulus principle.

Theorem 8.3.9 Suppose that f is a non-constant holomorphic function in an open setcontaining z0 and a disc D(z0, R). Then for each 0 < r < R, there exists a point inD(z0, r) s.t. |f(z)| > |f(z0)|.

Proof Firstly, f is not a constant on D(z0, r), otherwise f would be a constant every-where, c.f. Theorem 5.8.3. If f(z0) = 0 the conclusion holds ( there is a point z withf(z) 6= 0.) By Cauchy’s formula, for every z ∈ D(z0, r),

f(z0) =1

2πi

∫C(z0,r)

f(z)

z − z0dz =

1

2π

∫ 2π

0

f(z0 + eit)dt.

Suppose f(z0) 6= 0 and |f(z)| ≤ |f(z0)| on D(z0, r). Then |f(z)| ≤ f(z0)| onC(z0, r). We divide the equation above by f(z0) and split the real and pure imaginaryparts of the integrand: f(z)

f(z0) = u(z) + iv(z). Then u2(z) + v2(z) ≤ 1, and∫ 2π

0

u(z0 + eit)dt = 2π.

78

This can happen if and only if u is identically 1. (If u(z0) < 1 at one point thereexists a real number M < 1 s.t. f(z) ≤ M on an interval of non-zero length and∣∣∣∫ 2π

0u(z0 + eit)dt

∣∣∣ ≤ 2π.)Thus u ≡ 1 and v = 0 and f is a constant on the disc D(z0, r), a contradiction. We

conclude that there exists z ∈ D(z0, r) such that |f(z)| > |f(z0)|. �

8.4 Bi-holomorphic Maps on the Disc (lecture 25-26)We would like to understand how does a holomorphic function look like.

Lemma 8.4.1 (Schwartz’s Lemma) Let D = {z : |z| < 1}. Suppose that f is aholomorphic function on D, mapping D into D, and f(0) = 0. Then f satisfies one ofthe following descriptions:

• For all z, f(z) = cz for a complex number c with |c| = 1.

• |f(z)| < |z| for z with 0 < |z| < 1.

Proof Since f is holomorphic onD and f(0) = 0, we apply Lemma 5.8.1: there existsa holomorphic function h such that

f(z) = zh(z), z ∈ D.

Since |f(z)| ≤ 1, for any number 0 < r < 1,

1 ≥ sup|z|=r

|f(z)| = sup|z|=r

|z||h(z)| = r sup|z|=r

|h(z)|.

Thus sup|z|=r |h(z)| ≤ 1r for any r < 1. By the maximum modulus principle, Theorem

8.3.9,

sup|z|≤r

|h(z)| = sup|z|=r

|h(z)| ≤ 1

r.

Taking r → 1 we see that,sup|z|<1

|h(z)| ≤ 1.

If there exists z0 such that |h(z0)| = 1, then h has a local maximum at z0. Thush(z) = h(z0) a constant and f(z) = zh(z0).

Otherwise |g(z)| < 1 for any z ∈ D. Then for z 6= 0 and z ∈ D, |f(z)| =|z||g(z)| < |z|. �

Definition 8.4.1 A bijective holomorphic function f : U → V is said to be bi-holomorphic (In this case its inverse is also holomorphic.) We say U and V areconformally equivalent, if a bi-holomorphic map f : U → V exists.

Proposition 8.4.2 If f : D → D is a bi-holomorphic function with f(0) = 0. Thenf(z) = kz for some complex number k with |k| = 1.

79

Proof We apply Schwartz’s lemma to both f and f−1: |f(z)| ≤ |z| and |f−1(z)| ≤ |z|on D. Thus |f(z)| = |z| on D. By the same lemma, f(z) = kz for some complexnumber k. �

Denote Gb(z) = z−b1−bz the Mobius transform, where |b| < 1. Then Gb takes D

onto D, and (Gb)−1 = G−b. (See Example sheet 2)

Theorem 8.4.3 If f : D → D is a bi-holomorphic function, then there exist complexnumbers k, b with |k| = 1 and |b| < 1 such that f = kGb.

Proof There exists a unique point b such that f(b) = 0. Then f ◦ (Gb)−1 : D → D

is bi-holomorphic and takes 0 to 0. By Proposition 8.4.2, there exists k such thatf ◦ (Gb)

−1(z) = kz and f(z) = kGb(z). �

Remark 8.4.4 Let H := {z : Im(z) > 0}. Then f : H → H given by the formulaf(z) = az+b

cz+d , where a, b, c, d ∈ R with ad−bc = 1, are bi-holomorphic. (c.f. Examplesheet 2) All bi-holomorphic maps from H to H are of the above form. This can beproved by transferring our problem to D by the bi-holomorphic map F : H → D,F (z) = i+z

i−z . Its inverse is G(w) = i 1−w1+w .

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Chapter 9

The Riemann MappingTheorem

Given any two open sets, does there exist a bi-holomorphic function between them? IfU = C we know this is not possible if V is a bounded region, not a singleton. Suchmaps are bounded entire and so a constant by Liouville’s theorem. If f : U → V is bi-holomorphic (or a homeomorphism) and U is simply connected then V is also simplyconnected. If f : U → V is continuous, its image will be contained in a connectedsub-set of V . It is perhaps surprising that these properties determine all open sets thatare conformally equivalent to D. Recall that a region is an open connected subset ofC.

Theorem 9.0.1 (The Riemann Mapping Theorem) A simply connected regionU , whichis not the whole plane C, is conformally equivalent to the disc D = {z : |z| < 1}.

Main Idea. Let z0 ∈ U . Let

F = {f : U → D holomorphic, one to one, f(z0) = 0, f ′(z0) > 0}.

We prove that (1) F is not empty. (2) Prove B = sup{|f ′(z0)| : f ∈ F} <∞. Thereexist a sequence of functions fn such that limn→∞ f ′n(z0) = B. We prove that fn hasa subsequence, converging to a function f uniformly on compact subsets of U . (3) Iff ∈ F satisfies f ′(z0) = B, it is onto.

9.1 Hurwitz’s Theorem (lecture 26)

Example 9.1.1 Let fn(z) =∑nk=1

zk

k! . The entire functions fn are mapped onto C bythe Fundamental Theorem of Algebra. They converge uniformly on compact subsetsof C to ez . But ez does not take the value 0.

Example 9.1.2 let fn(z) = z0 + zn . These are entire bijective functions, converge

uniformly on compact subsets of C to the constant function z0, failing injectivity.

Theorem 9.1.3 (Hurwitz’s Theorem) Suppose fn is a sequence of holomorphic func-tions in a region U , converging to a function f uniformly on compact subsets of U .Suppose f is not a constant. Then the following statements hold.

81

1. Let w0 = f(z0) where z0 ∈ U . Then for sufficiently small ε, there exists anatural number N(ε) such that for each n ≥ N(ε), there exists zn ∈ D(z0, ε)such that fn(zn) = w0.

2. If each fn is one to one, so is f .

Proof By the uniform convergence theorem (Thm 5.5.1), f is holomorphic. Now,

fn(z)− w0 = (f(z)− w0) + (fn(z)− f(z)).

Since z0 is an isolated zero of f−w0, there exists a number ε > 0 such thatD(z0, 3ε) ⊂U and f(z) 6= w0 on D′(z0, 3ε). Let

δ = infz:|z−z0|=ε

|f(z)− w0| > 0.

Since fn converges uniformly to f on D(z0, 2ε), there exists N such that for n ≥ N ,|fn(z) − f(z)| < δ. For such n, we apply Rouche’s theorem: fn − w0 and f − w0

have the same number of zero’s inside D(z0, ε), in particular fn − w0 has at least onezero. This completes (1).

Suppose f(z1) = f(z2) for z1 6= z2. Denote w0 = f(z1). There exists ε >0 such that the discs D(z1, ε) and D(z2, ε) are disjoint and contained in U , and forn sufficiently large, w0 has a pre-image by fn in each of the discs. Such fn is notinjective, giving a contradiction, proving (2). �

9.2 Family of Holomorphic Functions (Lecture 28)If U is a region in C, we denote byH(U) the family of holomorphic functions on U .

Definition 9.2.1 Let U be a region in C and F a family of holomorphic functions onU . We say

1. F is normal or relatively compact, if every sequence from F has a sub-sequencethat converges uniformly on every compact subset of U .

2. F is locally uniformly bounded, if for any compact subset K of U , there existsM > 0 such that |f(z)| ≤M for any f ∈ F and z ∈ K.

3. F is equi-continuous on a compact set K of U if for every ε > 0 there existsδ > 0 such that if |z − w| < δ, z, w,∈ K,

|f(z)− f(w)| < ε, ∀, f ∈ F .

If there exists M such that |f ′| ≤ M for all f ∈ F then F is equi-continuous.The functions {fn(x) = xn : x ∈ [0, 1]} is not equi-continuous, nor has uniformconvergent subsequences (the limit is a discontinuous function).

Theorem 9.2.1 (Mantel’s Theorem) Suppose F is a locally uniformly bounded fam-ily of holomorphic functions in a region U , then F is normal.

The proof for Mantel’s theorem is split into two lemmas. The first one is specific toholomorphic functions, while the second is part of an Arzela-Ascoli theorem.

82

Lemma 9.2.2 Suppose F is a family of holomorphic functions uniformly bounded oncompact subsets of U , then F is equi-continuous on every compact subset of U .

Proof Let K be a compact subset of U . Then there exists a positive number r and anenlargement of K which is contained in U:

K = {z : the distance from z to K is less or equal to 4r} ⊂ U.

Let M = supf∈F,z∈K |f(z)|. Let z, w ∈ K with |z − w| < r. Then the circleγ = C(z, 2r) and its interior is contained in U .

f(z)− f(w) =1

2πi

∫γ

(f(ζ)

ζ − z− f(ζ)

ζ − w

)dζ =

1

2πi

∫γ

f(ζ)(z − w)

(ζ − z)(ζ − w)dζ.

For a given ε > 0, let δ = min( εr2M , r) and |z − w| < δ, z, w ∈ K,

|f(z)− f(w)| ≤ 4πr

2π{ supζ:|ζ−z|=2r}

|f(ζ)| 1

r2|z − w| ≤ 2M

r|z − w| < ε.

This proves f is equi-continuous on K. �

A sequence {Kl}∞l=1 of compact subsets of an open set U is an exhaustion if (1)Each Kl is contained in the interior of Kl+1, (2) Any compact subset K of U is con-tained in Kl for some l. Such an exhaustion always exists: take Kl = {z ∈ U : |z| ≤l, distance(z, ∂U) ≥ 1

l }.

Lemma 9.2.3 Suppose F is a locally uniformly bounded family of holomorphic func-tions in a region U which are equi-continuous on compact subsets of U , then F isnormal.

Proof Let {fn} be a sequence from F . Given a compact subset K of U let {wj} bea dense set of K. The sequence of numbers {fn(w1)} is bounded and hence has aconvergent subsequence given by a sub-sequence of function {fn,1}. From {fn,1} weextract another sequence {fn,2} such that fn,2(w2) converges as n → ∞. We repeatthis procedure and obtain nested sequence {{fn,k}, k = 1, 2, . . . , } with the propertythat for each k ≥ 1, fn,k(wk) converges. Let gn = fn,n be the diagonal sequence offunctions.

Since F is equi-continuous on K, for ε > 0 there exists δ > 0 such that |f(z) −f(w)| < ε

3 whenever |z − w| < δ and z, w ∈ K. Since K is compact we mayselect a finite number of points from {wj} which we denote by z1, . . . , zN such thatK ⊂ ∪Nk=1D(zj , δ). There is N(ε) > 0 such that for n,m ≥ N(ε),

|gn(zj)− gm(zj)| <ε

3, ∀j = 1, . . . , N .

If w ∈ K, there is a number j0 such that w ∈ D(zj0 , δ). Then

|gn(w)− gm(w)| ≤ |gn(w)− gn(zj0)|+ |gn(zj0)− gm(zj0)|+ |gm(zj0)− gm(w)|.

If n,m ≥ N(ε), observing |zj0 − w| < δ,

|gn(w)− gm(w)| ≤ ε

3+ε

3+ε

3.

Thus gn converges, uniformly on K.

83

We proceed to proceed that there is a sequence of gn which converges uniformlyon every compact set of U , by another diagonal argument. Let {Kn} be an exhaustionof U . Let {gn,1} be a subsequence of {gn} which converges uniformly on K1, {gn,2}a sub-sequence of {gn,1} converging uniformly on K2. Then the diagonal sequence{gn,n} converges uniformly on each Kn. Since every compact set is included in oneof the set from {Kl}, the proof is complete. �

9.3 The Riemann Mapping Theorem (Lecture 28-29)There are homeomorphisms (bijection, continuous with inverse continuous) from D toC, e.g. take f(z) = z

1−|z| ; there are no bi-holomorphic functions from D to C.

Theorem 9.3.1 (The Riemann Mapping Theorem) If U is a simply connected re-gion U , which is not the whole plane C, there exists a bi-holomorphic map from Uto the disc D = {z : |z| < 1}.

The bi-holomporphic maps from U to D is unique after we fix some parameters.

Proposition 9.3.2 Suppose U is a simply connected region which is not the whole C,and z0 ∈ U then there is at most one bi-holomorphic map f : U → D s.t. f(z0) = 0and f ′(z0) > 0.

Proof If f, g are two such maps, f ◦g−1 : D → D is holomorphic with f ◦g−1(0) = 0.Hence f = cg for a complex number c with |c| = 1. Now f ′(z0) = cg′(z0), bothderivatives are real numbers, then c is a real number and c = 1. �

Let us consider the following family of holomorphic functions:

F = {f : U → D holomorphic, one to one, f(z0) = 0}.

A function f on U is said to be non-vanishing if f(z) 6= 0 for any z ∈ U .

Lemma 9.3.3 If U is a simply connected and connected set which is not the wholeplane, there exists a bi-holomorphic function h : U → h(U) such that h(U) does notintersect a disc D(w0, δ) for some δ > 0 and some w0 ∈ C.

Proof Since U is a proper subset of C, there exists a ∈ C \ U . Then z − a is anon-vanishing holomorphic function on the simply connected region U . There exists aholomorphic function h : U → C such that z− a = eh(z) (see Theorem 8.0.1). Defineh(z) = e

12 h(z), a holomorphic function with

h2(z) = z − a.

We observe that h does not take the same value twice nor opposite values. (If h(z1) =±h(z2) then z1 − a = z2 − a.)

By the open mapping theorem, h(U) contains a disc D(h(z0), δ), it therefore can-not intersect D(−h(z0), δ), which contains opposite values. Let w0 = −h(z0). �

Lemma 9.3.4 F is not empty.

84

Proof By Lemma 9.3.3, we may assume that D(w0, δ) ∩ U = ∅ for some δ > 0,and w0 ∈ C and define f(z) = δ

z−w0for z ∈ U . Since |z − w0| > δ for all z ∈ U ,

f is holomorphic. It is clearly one to one, and |f(z)| < 1. If Gb : D → D is thebi-holomorphic map taking b to zero, Gb(z) = z−b

1−bz , then Gf(z0) ◦ f ∈ F . �

Since G′b(b) = 11−|b|2 , (Gf(z0) ◦ f)′(z0) = 1

1−|f(z0)|2 f′(z0) 6= 0.

Lemma 9.3.5 There exist f ∈ F s.t. |f ′(z0)| = supf∈F |f ′(z0)|.

Proof Take a disc D(z0, r) ⊂ U . By Cauchy’s inequality,

|f ′(z0)| ≤ 1

rsup

|z−z0|=r|f(z)| ≤ 1

r, ∀f ∈ F .

The set {|f ′(z0)|, f ∈ F} is bounded and has a supremum which we denote by B,B = supf∈F |f ′(z0)|. There exists a sequence fn ∈ F s.t. limn→∞ |f ′n(z0)| = B. ByMontel’s theorem, since |f | ≤ 1 for all f ∈ F , there exists a subsequence {fnk} of{fn} converging to a function f uniformly on compact subsets of U . By WeierstrassTheorem ( i.e. the Uniform Convergence Theorem, Theorem 5.5.1), f is holomorphic.Since |f ′(z0)| = limk→∞ |f ′nk(z0)| 6= 0, f is not a constant function and is thereforeone to one by Hurwitz’s theorem. Also f(z0) = limk→∞ fnk(z0) = 0. So f ∈ F . �

Let us review properties of the basic bi-holomorphic maps on D. For b ∈ Ddefine Gb(z) = z−b

1−bz , a bi-holomorphic function from D to D with Gb(z) = 0 and

Gb(0) = −b, G′b(z) = 1−|b|2(1−bz)2 , G′b(0) = 1− |b|2 and G′b(b) = 1

1−|b|2 .

Lemma 9.3.6 If f ∈ F , f is not onto, there exists g ∈ F s.t. |g′(z0)| > |f ′(z0)|.

Proof Let w0 ∈ D, w0 6∈ f(U). The Mobius transform Gw0has only one zero at

w0. Hence Gw0◦ f : U → D is one to one, holomorphic, and never vanishing. Since

Gw0 ◦ f is defined on a simply connected domain there exists a holomorphic functionh : U → C, a square root function, such that

(h(z))2 = Gw0◦ f(z) =

f(z)− w0

1− w0f(z), (9.3.1)

and (h(z0))2 = −w0. See the proof of Lemma 9.3.We now take h(z0) back to 0 by another mobius transform: g = Gh(z0) ◦ h. Then

g ∈ F and |g′(z0)| > |f ′(z0)|. To check the inequality holds, we first differentiate(9.3.1):

2h(z0)h′(z0) = Gw0(f(z0))f ′(z0) = (1− |w0|2)f ′(z0),

On the other hand,

|g′(z0)| = |G′h(z0)(h(z0)) · h′(z0)| = 1

1− |h(z0)|2(1− |w0|2)f ′(z0)

2|h(z0)|

= |f ′(z0)|1 + |w0|2√|w0|

< |f ′(z0)|.

This completes the proof. �

We have competed the proof for the Riemann mapping Theorem.Remark. Let U be an open connected set. The property that every a holomorphic

never vanishing function onU has a holomorphic square root characterises the propertythat U a simply connected.

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9.4 SupplementaryThe following concepts are equivalent for an open connected subset U of C.

1. U is simply connected;

2. C∗ \ U is connected.

3. Every holomorphic function on U has a primitive;

4. Any non-vanishing holomorphic function on U has a holomorphic square root.

5. Any non-vanishing holomorphic function on U has a holomorphic logarithm, i.e.a holomorphic function h such that eh = f .

6. U is homeomorphic to a disc.

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Chapter 10

Special Functions

10.1 Constructing Holomorphic Functions by Integra-tion (Lecture 30)

The following theorem is a simple corollary of Mantel’s Theorem. We will give insteadan elementary proof for it.

Theorem 10.1.1 Let U be a region in C and G : U × [0, 1]→ C a continuous functions.t. G(·, s) is holomorphic for each s ∈ [0, 1]. Then the function f(z) =

∫ 1

0G(z, s)ds

is holomorphic.

Proof We define a family of holomorphic functions:

fn(z) =

n∑k=1

∫ kn

k−1n

G(z,k

n)ds =

1

n

n∑k=1

G(z,k

n).

Let z0 ∈ U andD(z0, 2r) a disc in U . ThenG is uniformly continuous on the compactset D(z0, r). Indeed, for any ε > 0, there exists δ > 0 such that if |s − t| < δ,|G(z, s)−G(z, t)| < ε for all z ∈ D(z0, r). If n > 1

δ , and z ∈ D(z0, r)

|fn(z)− f(z)| =

∣∣∣∣∣n∑k=1

∫ kn

k−1n

(G(z,k

n)−G(z, s))ds

∣∣∣∣∣ < ε.

Hence fn → f converges uniformly on D(z0, r). By Theorem 5.5.1, f is holomorphicon D(z0, r). Since z0 is an arbitrary point of U , f is holomorphic in U . �

10.2 The Gamma functionA complex differentiable (holomorphic functions) has a power series expansion aroundevery point (analytic). In this section we switch terminology and use analytic for holo-morphic.

For any positive real number s, the Gamma function

Γ(s) =

∫ ∞0

e−tts−1dt (10.2.1)

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is well defined. At 0, ts−1 is integrable for s > 1. Let Γε(s) =∫ 1ε

εe−tts−1dt. For

t large e−t dominates and is integrable. Hence Γ(ε) converges if s is real and greaterthan 1.

Lemma 10.2.1 Formula (10.2.1) defines an analytic function on {s : Re(s) > 0}.

Proof Firstly |ts−1| = |e(Re(s)+iIm(s)−1) ln t = tRe(s)−1. Let

Γε =

∫ 1ε

ε

e−tts−1dt,

each of which is holomorphic by Theorem 10.1.1. They converge uniformly on thestrips An = {s : 1

n < Re(s) < n}. (Check that |Γε(s) − Γ(s)| → 0 uniformly asε→ 0.) Thus Γ is an analytic function. �

Lemma 10.2.2 If Re(s) > 0, Γ(s + 1) = sΓ(s). Consequently Γ(n + 1) = n! forn = 0, 1, 2, . . . .

Proof For any ε > 0,∫ 1ε

ε

d

dt(e−tts)dt = −

∫ 1ε

ε

e−ttsdt+ s

∫ 1ε

ε

e−tts−1dt.

But limε→0

∫ 1ε

εddt (e

−tts)dt = limε→0

(e−

1ε ( 1ε )s − e−εεs

)= 0. The right hand side

converges to −Γs+1 + sΓ(s). �

Theorem 10.2.3 The function Γ extends to a meromorphic function into C whose onlysingularity are simple poles at s = 0,−1,−2, . . . .

Proof On {s : Re(s) > −1}, we define

Γ1(s) =Γ(s+ 1)

s.

Then Γ1 agree with Γ for s > 1, and has a simple pole at s = 0. Indeed, sinceΓ(1) = 1, lims→0 Γ1(s) = lims→0

Γ(s+1)s =∞ proving 0 is a pole.

On {s : Re(s) > −2} we define

Γ2(s) =Γ1(s+ 1)

s=

Γ(s+ 2)

s(s+ 1).

Inductively we define on {s : Re(s) > −n}

Γn(s) =Γn−1(s+ 1)

s=

Γ(s+ n)

s(s+ 1) . . . (s+ n− 1),

where Γ(s+n) is holomorphic on {s : Re(s) > −n}. It is clear that Γn(s) has simplepoles at 0,−1, . . . ,−(n − 1), also Γn agree with Γn+1 where they overlap. Hence Γhas a continuation given by the above formula on each strip {s : Re(s) > −n}. �

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It is easy to compute the residues: Res(Γ,−n) = (−1)n

n! . Also, 1Γ(s) is an en-

tire function whose only zero’s are simple zeros at 0,−1,−2, . . . , following from theidentity

Γ(s)Γ(1− s) =π

sin(πs), ∀s ∈ C.

In other words, 1Γ(s) = Γ(1−s) π

sin(πs) . The simple poles of Γ(1−s), at s = 1, 2, 3, . . .

and the simple zero’s of sin(πs) cancel.

10.3 The zeta function(Lecture 30)Let Re(s) > 1, the following infinite sum converges

ζ(s) =

∞∑n=1

1

ns

and we have an analytic function on the strip {s : Re(s) > 1}.

Theorem 10.3.1 The zeta function has a meromorphic continuation to C whose onlysingularity is a simple pole at s = 1.

Idea of the Proof.

1. Define the theta function θ(t) =∑∞n=−∞ e−πn

2t for t > 0. It has the followingproperties: |θ(t)− 1| ≤ ce−πt for t ≥ 1 and |θ(t)| ≤ Ct− 1

2 as t→ 0.Define the xi function:

ξ(s) =1

2

∫ ∞0

rs2−1(θ(r)− 1)dr.

Then ξ is holomorphic for Re(s) > 1 and has an meromorphic extension to Cwith simple poles at 0 and −1, Moreover ξ(s) = ξ(1− s) for all s ∈ C.

2. The following holds for Re(s) > 1,

ζ(s) =ξ(s)π

s2

Γ(s/2).

Since ξ has an meromorphic extension to C whose only singularity are simplepoles at 0 and 1, ζ has a meromorphic extension to C with a simple pole at 1.The simple pole of ξ(s) at s = 0 cancels with the simple zero of Γ(s/2) at s = 0.

The Euler’s identity states:

ζ(s) = Πp1

1− p−s

where the product is taken over all prime number. The heuristic argument for a proofthat there are an infinite number of prime numbers is as following. If we take s = 0,we see

∑∞n=1

1n = Πp

11−p−1 . Since the harmonic series diverges, there must be an

infinite number of terms on the right hand side.Euler’s trick is to consider both sides as functions on the complex plane.

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Theorem 10.3.2 There are an infinite number of prime numbers.

Proof If there are only a finite number of prime numbers, then x := Πp1

1−p−1 > 0 isa finite number. But by monotonicity, for any s > 1,

ζ(s) = Πp1

1− p−s< Πp1− p−1 = x.

But lims→1 ζ(s) =∞ since it has a pole at 1, giving a contradiction.�

Note that ζ(s) has simple zeros at s = −2,−4, . . . , inherited from 1Γ(s/2) . In fact

it is fairly easy to see ζ(s) does not vanish if Re(s) > 1.

ζ(s) = πs−1/2 Γ( 1−s2 )

Γ( s2 )ζ(1− s).

Let us consider the region Re(s) < 0 where Re(1− s) > 1 and ζ(1− s) has no zero,Γ( 1−s

2 ) has no zero, 1Γ( s2 ) has zero at 0,−2,−4, . . . . From this we conclude that

Theorem 10.3.3 The only zero’s of the zeta function outside of the critical strip {s : 0 ≤ Re(s) ≤ 1}are at the negative even integers −2,−4,−6, . . . . Furthermore, there are no zeros onthe vertical line Re(s) = 1.

The negative even integers are called the trivial zero’s of the zeta function.

Riemann’s Hypothesis. The zeros of the zeta function in the critical strip {s : 0 ≤ Re(s) ≤ 1}lie on the vertical line Re(s) = 1

2 .

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